Linked List – Computer Science

Do My Computer Homework!

1)      Assume you are working with a doubly linked list, based on the following node:

class node {

float data;

node next;

nodeprev;

}

Write an insert module – which takes a “head” pointer as a parameter (along with the data to be inserted), and inserts to the head of the list.  Consider the empty list case.

 

public void insert(float float_var){

            Node last = post.prev;

            Node x = new Node();

            x.data = float_var;

            x.next = last;

            x.prev = post;

            post.next = x;

            last.prev = x;

}

 

2)      Given the declarations below, write a recursive isThere module for a linked list implementation of the “Node” class.  The method will take 2 values – a “reference” to the node being looked at called “pointer”, and a search value called “lookingfor”.  Assume the existence of a boolen isEmpty method.  Also assume the list is “capped off” with Null.

class node {

int data;

node next;

}

                        public Boolean isThere(node reference, int value){

                                    node test = reference;

                                    if(!isEmpty(test)){

                                                if(test.data == value)

                                                            return true;

                                                else

                                                            isThere(test.next, value);

                                    }

                                    else

                                                return false;

}

3)      I have a stack implemented with an array (20 cells) of string objects. There is an integer variable called Top which holds the index of first open cell on the stack (the top) (example shown below).Write out the pseudocode for a “pop” method.  Create an “is_empty” method which returns a Boolean value (true for empty, false for not empty), then call it from your pop method.

 

[3]

[2]

3

[1]

Stack                 [0]                  Top

Is_empty method :

public Boolean is_Empty(int[] arrayStack, index_level){

            if (arrayStack[index_level] != NULL)                 

//Check if the Stack at index level is empty

                        return false;

            return true;

}

 

Pseudocode for pop method :

Function pop is

if Top is greater than zero

            set arrayStack at index Top := NULL

            Subtract one from Top 

 

4)      I have a queue implemented with a linked list.  There are two variables (front and rear).  Write the pseducode for a “count” method.  This method will return the number of items in the queue.  Assume that this data is not stored anywhere else.

 

Pseudocode for count method :

 

Function count(node) is     // takes any node within the queue

            Queue_length := 1

                                                Set frontpointer := pointer to front

            set rearpointer := pointer to rear

                                        //Split the nodes and start counting

            while frontpointer is not NULL               // Front Counting

                        Queue_length++

                        Set frontpointer := pointer to front

            while rearpointer is not NULL                 // Rear Counting

                        Queue_length++

                        Set rearpointer := pointer to rear

            Return Queue_length;

 

5)      Write the pseducode for a method that adds new items to the “end” of a linked list (list nodes hold a character as data).  The list only has a “head” pointer, so your method needs to make its way through the list to add the new element.  Assume the list is capped off with NULL.  Watch for the empty case.

 

Pseducode:

Function add_end(list, pointer to head, value) is

            While pointer to head is not NULL           // traverse till the end of the list

                        Set Pointer to head := list.next

            Initialize node new_node := new node  // create a new node

            Set list.next := pointer to new_node        // assign next pointer to the                                                                                               new_node

            Set new_node.data:= value                       // assign values to the new node

            Set new_node.next:= NULL                      // next points to null because it                                                                                        is at the tail of the list

 

 

 

Solving Equation II

f(x)=3x-4,g(x)=x+2

 

Setup the composite result function.  The fog notation is interpreted as f(g(x)).

f(g(x))

 

Evaluate f(g(x)) by substituting in the value of g into f.

f(x+2)=3(x+2)-4

 

Multiply 3 by each term inside the parentheses.

f(x+2)=3x+6-4

 

Subtract 4 from 6 to get 2.

f(x+2)=3x+2

 

(G*F)= f(x)=3x-4,g(x)=x+2

 

Multiply f by each term inside the parentheses.

f(x)=3x-4_g(x)=x+2_g(fx)

 

Multiply g by each term inside the parentheses.

f(x)=3x-4_g(x)=x+2_fgx

 

Compose the result function for fgx by replacing the function designators with the actual functions.

(3x-4)(x+2)x

 

Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group.

x(3x*x+3x*2-4*x-4*2)

 

Simplify the FOIL expression by multiplying and combining all like terms.

x(3x^(2)+2x-8)

 

Multiply x by each term inside the parentheses.

3x^(3)+2x^(2)-8x

 

Now (F*G)(2)=3*2+2 = 8

 

 

2.

F(x)=√x  and g(x)=x-1

 

f(x)=√(x)

 

The domain of an expression is all real numbers except for the regions where the expression is undefined.  This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0.  All of these are undefined and therefore are not part of the domain.

(x)<0

 

Solve the equation to find where the original expression is undefined.

x<0

 

The domain of the rational expression is all real numbers except where the expression is undefined.

[0, ,∞)

 

f(x)=x-1

 

The domain of the rational expression is all real numbers except where the expression is undefined.  In this case, there is no real number that makes the expression undefined.

All real numbers

 

 

3.

F(x)= f(x)=√(84-6x)

 

The domain of an expression is all real numbers except for the regions where the expression is undefined.  This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0.  All of these are undefined and therefore are not part of the domain.

(84-6x)<0

 

Solve the equation to find where the original expression is undefined.

x=14

 

The domain of the rational expression is all real numbers except where the expression is undefined.

(-∞,14) U (14,∞)

 

4.

F(x)=x^3+2

f(x)=x^(3)+2

 

To find the inverse of the function, interchange the variables and solve for f^(-1)(x).

x=f^(-1)(x)^(3)+2

 

Since f^(-1)(x) is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

f^(-1)(x)^(3)+2=x

 

Since 2 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 2 from both sides.

f^(-1)(x)^(3)=-2+x

 

Move all terms not containing f^(-1)(x) to the right-hand side of the equation.

f^(-1)(x)^(3)=x-2

 

Take the cube root of both sides of the equation to eliminate the exponent on the left-hand side.

 

 

f^(-1)(x)=√3:(x-2)

 

 

 

 

1.

Distance between points

(5,1),(8,5)

 

Use the distance formula to determine the distance between the two points.

distance=√((x2-x1)^(2)+(y2-y1)^(2))

 

Substitute the actual values of the points into the distance formula.

distance=√((8-(5))^(2)+(5-(1))^(2))

 

Multiply -1 by the 5 inside the parentheses.

distance=√((8-5)^(2)+(5-(1))^(2))

 

Subtract 5 from 8 to get 3.

distance=√((3)^(2)+(5-(1))^(2))

 

Expand the exponent (2) to the expression.

distance=√((3^(2))+(5-(1))^(2))

 

Squaring a number is the same as multiplying the number by itself (3*3).  In this case, 3 squared is 9.

distance=√((9)+(5-(1))^(2))

 

Multiply -1 by the 1 inside the parentheses.

distance=√((9)+(5-1)^(2))

 

Subtract 1 from 5 to get 4.

distance=√((9)+(4)^(2))

 

Expand the exponent (2) to the expression.

distance=√((9)+(4^(2)))

 

Squaring a number is the same as multiplying the number by itself (4*4).  In this case, 4 squared is 16.

distance=√((9)+(16))

 

Remove the parentheses that are not needed from the expression.

distance=√(9+16)

 

Add 16 to 9 to get 25.

distance=√(25)

 

Pull all perfect square roots out from under the radical.  In this case, remove the 5 because it is a perfect square.

distance=5 units

 

 

(10,4) and (2,6)

 

(10,4),(2,6)

 

Use the midpoint formula to find the midpoint of the line segment.

((x1+x2)/(2) , (y1+y2)/(2))

 

Substitute in the values for (x1,y1) and (x2,y2).

((10+2)/(2),(4+6)/(2))

 

Add 2 to 10 to get 12.

((12)/(2),(4+6)/(2))

 

Reduce the expression (12)/(2) by removing a factor of 2 from the numerator and denominator.

(6,(4+6)/(2))

 

Add 6 to 4 to get 10.

(6,(10)/(2))

 

Reduce the expression (10)/(2) by removing a factor of 2 from the numerator and denominator.

(6,5)

 

3.

Equation OF Circle

 

(x-(-3))^2+(y-5)^2=3^2

(x+9)^2+(y-5)^2=9

 

X^2+y^2 + 18x -10y + 97=0

 

 

4.

X^2+y^2+8x+4y+16=0

(X+4)^2 + Y^2 + 4Y + 4 – 4 =0

(X+4)^2+(Y+2)^2 = 2^2

 

 

Radius = 2 units

 

 

8.

 

f(x)=2x^(2)-7x-4

 

Replace f(x) with y to find the properties of the parabola.

y=2x^(2)-7x-4

 

To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of x.  In this problem, add (-(7)/(4))^(2) to both sides of the equation.

y=2(x^(2)-(7x)/(2)+(49)/(16))+2(-2)-(2)(0+(49)/(16))

 

Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression.

y=2(x^(2)-(7x)/(2)+(49)/(16))+2(-2)-(2)((49)/(16))

 

Factor the perfect trinomial square into (x-(7)/(4))^(2).

y=2((x-(7)/(4))^(2))+2(-2)-(2)((49)/(16))

 

Factor the perfect trinomial square into (x-(7)/(4))^(2).

y=2(x-(7)/(4))^(2)+2(-2)-(2)((49)/(16))

 

Multiply 2 by each term inside the parentheses.

y=2(x-(7)/(4))^(2)-4-(2)((49)/(16))

 

Multiply 2 by (49)/(16) to get (49)/(8).

y=2(x-(7)/(4))^(2)-4-((49)/(8))

 

Multiply -1 by the (49)/(8) inside the parentheses.

y=2(x-(7)/(4))^(2)-4-(49)/(8)

 

To add fractions, the denominators must be equal.  The denominators can be made equal by finding the least common denominator (LCD).  In this case, the LCD is 8.  Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.

y=2(x-(7)/(4))^(2)-4*(8)/(8)-(49)/(8)

 

Complete the multiplication to produce a denominator of 8 in each expression.

y=2(x-(7)/(4))^(2)-(32)/(8)-(49)/(8)

 

Combine the numerators of all fractions that have common denominators.

y=2(x-(7)/(4))^(2)+(-32-49)/(8)

 

Subtract 49 from -32 to get -81.

y=2(x-(7)/(4))^(2)+(-81)/(8)

 

Move the minus sign from the numerator to the front of the expression.

y=2(x-(7)/(4))^(2)-(81)/(8)

 

This is the form of a paraboloa.  Use this form to determine the values used to find vertex and x-y intercepts.

y=a(x-h)^(2)+k

 

Use the standard form to determine the vertex and x-y intercepts.

a=2_k=-(81)/(8)_h=(7)/(4)

 

The vertex of a parabola is (h,k).

Vertex: ((7)/(4),-(81)/(8))

 

Domainf(x)=2x^(2)-7x-4

 

The domain of the rational expression is all real numbers except where the expression is undefined.  In this case, there is no real number that makes the expression undefined.

All real numbers

f(x)=2x^(2)-7x-4

 

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

2x^(2)-7x-4=f(x)

 

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.

2x^(2)-7x-f(x)-4=0

 

Use the quadratic formula to find the solutions.  In this case, the values are a=2, b=-7, and c=-1f(x)-4.

x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

 

Use the standard form of the equation to find a, b, and c for this quadratic.

a=2, b=-7, and c=-1f(x)-4

 

Substitute in the values of a=2, b=-7, and c=-1f(x)-4.

x=(-(-7)\~((-7)^(2)-4(2)(-1f(x)-4)))/(2(2))

 

Multiply -1 by each term inside the parentheses.

x=(7\~((-7)^(2)-4(2)(-1f(x)-4)))/(2(2))

 

Simplify the section inside the radical.

x=(7\~(8f(x)+81))/(2(2))

 

Simplify the denominator of the quadratic formula.

x=(7\~(8f(x)+81))/(4)

 

First, solve the + portion of \.

x=(7+~(8f(x)+81))/(4)

 

Next, solve the – portion of \.

x=(7-~(8f(x)+81))/(4)

 

The final answer is the combination of both solutions.

x=(7+~(8f(x)+81))/(4),(7-~(8f(x)+81))/(4)

 

The domain of an expression is all real numbers except for the regions where the expression is undefined.  This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0.  All of these are undefined and therefore are not part of the domain.

(8f(x)+81)<0

 

Solve the equation to find where the original expression is undefined.

f(x)<-(81)/(8)

 

The domain of the rational expression is all real numbers except where the expression is undefined.

f(x)>=-(81)/(8)_[-(81)/(8),infinity

 

The domain of the inverse of f(x)=2x^(2)-7x-4 is equal to the range of f(f(x))=((7+~(8f(x)+81)))/(4).

Range: f(x)>=-(81)/(8)

=  [-(81)/(8),infinity]

 

 

 

2.

F(x) = 6- 4x+ x^2

f(x)=6-4x+x^(2)

 

Replace f(x) with y to find the properties of the parabola.

y=6-4x+x^(2)

 

Nothing

y=-4(-(x^(2))/(4)+x)

 

Nothing

y=4(-(x^(2))/(4)*-4+x*-4)-(-4)(0*-4)

 

 

Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression.

y=-4(x^(2)-4x+4)-(-4)(4)

 

Factor the perfect trinomial square into (x-2)^(2).

y=-4((x-2)^(2))-(-4)(4)

Factor the perfect trinomial square into (x-2)^(2).

y=-4(x-2)^(2)-(-4)(4)

 

Multiply -4 by 4 to get -16.

y=-4(x-2)^(2)-(-16)

 

Multiply -1 by each term inside the parentheses.

y=-4(x-2)^(2)+16

 

This is the form of a paraboloa.  Use this form to determine the values used to find vertex and x-y intercepts.

y=a(x-h)^(2)+k

 

Use the standard form to determine the vertex and x-y intercepts.

a=-4_k=16_h=2

 

The vertex of a parabola is (h,k).

Vertex: (2,16)

 

 

 

 

Graph

f(x)=6-4x+x^(2)

 

The domain of the rational expression is all real numbers except where the expression is undefined.  In this case, there is no real number that makes the expression undefined.

All real numbers

f(x)=6-4x+x^(2)

 

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

6-4x+x^(2)=f(x)

 

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.

x^(2)-4x-f(x)+6=0

 

Use the quadratic formula to find the solutions.  In this case, the values are a=1, b=-4, and c=-1f(x)+6.

x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

 

Use the standard form of the equation to find a, b, and c for this quadratic.

a=1, b=-4, and c=-1f(x)+6

 

Substitute in the values of a=1, b=-4, and c=-1f(x)+6.

x=(-(-4)\~((-4)^(2)-4(1)(-1f(x)+6)))/(2(1))

 

Multiply -1 by each term inside the parentheses.

x=(4\~((-4)^(2)-4(1)(-1f(x)+6)))/(2(1))

 

Simplify the section inside the radical.

x=(4\2~(-1(-f(x)+2)))/(2(1))

 

Simplify the denominator of the quadratic formula.

x=(4\2~(-1(-f(x)+2)))/(2)

 

First, solve the + portion of \.

x=(4+2~(-1(-f(x)+2)))/(2)

 

Simplify the expression to solve for the + portion of the \.

x=2+~(-1(-f(x)+2))

 

Next, solve the – portion of \.

x=(4-2~(-1(-f(x)+2)))/(2)

 

Simplify the expression to solve for the – portion of the \.

x=2-~(-1(-f(x)+2))

 

The final answer is the combination of both solutions.

x=2+~(-1(-f(x)+2)),2-~(-1(-f(x)+2))

 

The domain of an expression is all real numbers except for the regions where the expression is undefined.  This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0.  All of these are undefined and therefore are not part of the domain.

(-1(-f(x)+2))<0

 

Solve the equation to find where the original expression is undefined.

f(x)<2

 

The domain of the rational expression is all real numbers except where the expression is undefined.

f(x)>=2_[2,infinity

 

The domain of the inverse of f(x)=6-4x+x^(2) is equal to the range of f(f(x))=2+~(-1(-f(x)+2)).

Range: f(x)>=2_[2,infinity)

 

3.

 

f(x)=-3*(x+(1)/(2))(x-4)^(3)

 

To find the roots)/(zeros of the function, set the function equal to 0 and solve.

0=-3*(x+(1)/(2))(x-4)^(3)

 

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

-3*(x+(1)/(2))(x-4)^(3)=0

 

Multiply -3 by (x+(1)/(2))(x-4)^(3) to get -3(x+(1)/(2))(x-4)^(3).

-3(x+(1)/(2))(x-4)^(3)=0

 

If any individual factor on the left-hand side of the equation is equal to 0, the entire expression will be equal to 0.

(x+(1)/(2))=0_(x-4)^(3)=0

 

Set the first factor equal to 0 and solve.

(x+(1)/(2))=0

 

Remove the parentheses around the expression x+(1)/(2).

x+(1)/(2)=0

 

Since (1)/(2) does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting (1)/(2) from both sides.

x=-(1)/(2)

 

Set the next factor equal to 0 and solve.

(x-4)^(3)=0

 

Set each of the factors of the left-hand side of the equation equal to 0.

x-4=0

 

Since -4 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 4 to both sides.

x=4

 

The final solution is all the values that make -3(x+(1)/(2))(x-4)^(3)=0 true.  The multiplicity of a root is the number of times the root appears.  For example, a factor of (x-4)^(3)=0 has multiplicity of 3.

x=-(1)/(2),4 with multiplicity: 3

 

 

the graph  crosses the x axis

 

 

4.

Leading co-efficient test

F(x)=-2*x^4+2*x^3

When the leading coefficient is negative and the degree is even, it falls to the left and falls to the right

 

f(F,x)=-2*x^(4)+2*x^(3)

 

To find the x-intercept, substitute in 0 for y and solve for x.

(0)=-2*x^(4)+2*x^(3)

 

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

-2*x^(4)+2*x^(3)=(0)

 

Multiply -2 by x^(4) to get -2x^(4).

-2x^(4)+2*x^(3)=(0)

 

Multiply 2 by x^(3) to get 2x^(3).

-2x^(4)+2x^(3)=(0)

 

Remove the parentheses around the expression 0.

-2x^(4)+2x^(3)=0

 

Multiply each term in the equation by -1.

2x^(4)-2x^(3)=0

 

Factor out the GCF of 2x^(3) from each term in the polynomial.

2x^(3)(x)+2x^(3)(-1)=0

 

Factor out the GCF of 2x^(3) from 2x^(4)-2x^(3).

2x^(3)(x-1)=0

 

Set the single term factor on the left-hand side of the equation equal to 0.

2x^(3)=0

 

Divide each term in the equation by 2.

(2x^(3))/(2)=(0)/(2)

 

Simplify the left-hand side of the equation by canceling the common factors.

x^(3)=(0)/(2)

 

0 divided by any number or variable is 0.

x^(3)=0

 

Take the cube root of both sides of the equation to eliminate the exponent on the left-hand side.

x=cubic root(0)

 

Pull all perfect cube roots out from under the radical.  In this case, remove the 0 because it is a perfect cube.

x=0

 

Set each of the factors of the left-hand side of the equation equal to 0.

x-1=0

 

Since -1 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 1 to both sides.

x=1

 

The complete solution is the set of the individual solutions.

x=0,1

 

To find the y-intercept, substitute in 0 for x and solve for y.

y=-2*(0)^(4)+2*(0)^(3)

 

Expand the exponent (4) to the expression.

y=-2*(0^(4))+2*(0)^(3)

 

Raising a number to the 4th power is the same as multiplying the number by itself 4 times.  In this case, 0 raised to the 4th power is 0.

y=-2*(0)+2*(0)^(3)

 

Expand the exponent (3) to the expression.

y=-2*(0)+2*(0^(3))

 

Cubing a number is the same as multiplying the number by itself 3 times (0*0*0).  In this case, 0 cubed is 0.

y=-2*(0)+2*(0)

 

Multiply all the factors separated by a * in -2*(0)+2*(0).

y=-2(0)+2(0)

 

Multiply -2 by each term inside the parentheses.

y=0+2(0)

 

Multiply 2 by each term inside the parentheses.

y=0+0

 

Solve the equation.

y=0

 

These are the x and y intercepts of the equation y=-2*x^(4)+2*x^(3).

x=0,1, y=0

 

 

 

 

We can see the graph has no symmetry

 

 

Solving Equations

1.

(x+2)^(2)=-7

 

Take the square root of each side of the equation to setup the solution for x.

√((x+2)^(2))=\√(-7)

 

Remove the perfect root factor (x+2) under the radical to solve for x.

(x+2)=\√(-7)

 

Pull all perfect square roots out from under the radical.  In this case, remove the i because it is a perfect square.

(x+2)=\i√(7)

 

First, substitute in the + portion of the \ to find the first solution.

(x+2)=i√(7)

 

Remove the parentheses around the expression x+2.

x+2=i√(7)

 

Since 2 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 2 from both sides.

x=-2+i√(7)

 

Move all terms not containing x to the right-hand side of the equation.

x=i√(7)-2

 

Next, substitute in the – portion of the \ to find the second solution.

(x+2)=-i√(7)

 

Remove the parentheses around the expression x+2.

x+2=-i√(7)

 

Since 2 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 2 from both sides.

x=-2+-i√(7)

 

Move all terms not containing x to the right-hand side of the equation.

x=-i√(7)-2

 

The complete solution is the result of both the + and – portions of the solution.

x=i√(7)-2,-i√(7)-2

 

 

 

2.

x^(2)+5x+2=0

 

Use the quadratic formula to find the solutions.  In this case, the values are a=1, b=5, and c=2.

x=(-b\√(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

 

Use the standard form of the equation to find a, b, and c for this quadratic.

a=1, b=5, and c=2

 

Substitute in the values of a=1, b=5, and c=2.

x=(-5\√((5)^(2)-4(1)(2)))/(2(1))

 

Simplify the section inside the radical.

x=(-5\√(17))/(2(1))

 

Simplify the denominator of the quadratic formula.

x=(-5\√(17))/(2)

 

First, solve the + portion of \.

x=(-5+√(17))/(2)

 

Next, solve the – portion of \.

x=(-5-√(17))/(2)

 

The final answer is the combination of both solutions.

x=(-5+√(17))/(2),(-5-√(17))/(2)_x<Z>APPR<z>-0.44,-4.56

 

 

3.

x+1-9x^(3)-9x^(2)=0

 

Multiply each term in the equation by -1.

9x^(3)+9x^(2)-x-1=0

 

Factor the greatest common factor (GCF) from each group.

(9x^(2)(x+1)-1(x+1))=0

 

Factor the polynomial by grouping the first two terms together and finding the greatest common factor (GCF).  Next, group the second two terms together and find the GCF.  Since both groups contain the factor (x+1), they can be combined.

(9x^(2)-1)(x+1)=0

 

The binomial can be factored using the difference of squares formula, because both terms are perfect squares.  The difference of squares formula is a^(2)-b^(2)=(a-b)(a+b).

(3x-1)(3x+1)(x+1)=0

 

Set each of the factors of the left-hand side of the equation equal to 0.

3x-1=0_3x+1=0_x+1=0

 

Since -1 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 1 to both sides.

3x=1_3x+1=0_x+1=0

 

Divide each term in the equation by 3.

(3x)/(3)=(1)/(3)_3x+1=0_x+1=0

 

Simplify the left-hand side of the equation by canceling the common factors.

x=(1)/(3)_3x+1=0_x+1=0

 

Set each of the factors of the left-hand side of the equation equal to 0.

x=(1)/(3)_3x+1=0_x+1=0

 

Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 1 from both sides.

x=(1)/(3)_3x=-1_x+1=0

 

Divide each term in the equation by 3.

x=(1)/(3)_(3x)/(3)=-(1)/(3)_x+1=0

 

Simplify the left-hand side of the equation by canceling the common factors.

x=(1)/(3)_x=-(1)/(3)_x+1=0

 

Set each of the factors of the left-hand side of the equation equal to 0.

x=(1)/(3)_x=-(1)/(3)_x+1=0

 

Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 1 from both sides.

x=(1)/(3)_x=-(1)/(3)_x=-1

 

The complete solution is the set of the individual solutions.

x=(1)/(3),-(1)/(3),-1

 

 

4.

√(x+10)=x-2

 

To remove the radical on the left-hand side of the equation, square both sides of the equation.

(√(x+10))^(2)=(x-2)^(2)

 

Simplify the left-hand side of the equation.

x+10=(x-2)^(2)

 

Squaring an expression is the same as multiplying the expression by itself 2 times.

x+10=(x-2)(x-2)

 

Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group.

x+10=(x*x+x*-2-2*x-2*-2)

 

Simplify the FOIL expression by multiplying and combining all like terms.

x+10=(x^(2)-4x+4)

 

Remove the parentheses around the expression x^(2)-4x+4.

x+10=x^(2)-4x+4

 

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

x^(2)-4x+4=x+10

 

Since x contains the variable to solve for, move it to the left-hand side of the equation by subtracting x from both sides.

x^(2)-4x+4-x=10

 

Since -4x and -x are like terms, subtract x from -4x to get -5x.

x^(2)-5x+4=10

 

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.

x^(2)-5x-6=0

 

In this problem 1*-6=-6 and 1-6=-5, so insert 1 as the right hand term of one factor and -6 as the right-hand term of the other factor.

(x+1)(x-6)=0

 

Set each of the factors of the left-hand side of the equation equal to 0.

x+1=0_x-6=0

 

Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 1 from both sides.

x=-1_x-6=0

 

Set each of the factors of the left-hand side of the equation equal to 0.

x=-1_x-6=0√

 

Since -6 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 6 to both sides.

x=-1_x=6

 

The complete solution is the set of the individual solutions.

x=-1,6

 

Verify each of the solutions by substituting them back into the original equation √(x+10)=x-2 and solving.  In this case, the solution (-1) was proven to be invalid during this process.

x=6

 

5.

 

Let x^1/3  =  z

So we substitute

2z^2 +7z- 15 = 0

2z^2+ 10 z – 3z -15 =0

2z(z+5) – 3(z+5)= 0

(2z-3)(z+5)=0

Z= 3/2,-5

X^1/3=3/2

x = 27/8  , -125

 

6.

7-(4)/(5)*x<(3)/(5)

 

Multiply -(4)/(5) by x to get -(4x)/(5).

7-(4x)/(5)<(3)/(5)

 

Since 7 does not contain the variable to solve for, move it to the right-hand side of the inequality by subtracting 7 from both sides.

-(4x)/(5)<-7+(3)/(5)

 

Simplify the right-hand side of the inequality.

-(4x)/(5)<-(32)/(5)

 

Multiply each term in the inequality by 5.

-(4x)/(5)*5<-(32)/(5)*5

 

Simplify the left-hand side of the inequality by canceling the common factors.

-4x<-(32)/(5)*5

 

Simplify the right-hand side of the inequality by simplifying each term.

-4x<-32

 

Divide each term in the inequality by -4.  When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

-(4x)/(-4)>-(32)/(-4)

 

Simplify the left-hand side of the inequality by canceling the common factors.

x>-(32)/(-4)

 

Simplify the right-hand side of the inequality by simplifying each term.

x>8

 

 

7.

7<x+5<11

 

Move all terms not containing x from the center section of the interval inequality.

7-5<x<11-5

 

Subtract 5 from 7 to get 2.

2<x<11-5

 

Subtract 5 from 11 to get 6.

2<x<6

 

Convert the solution to set notation.

(2,6)

 

8

 

-2|5-x|<-6

 

-2*|5-x|<-6

 

Multiply -2 by |5-x| to get -2|5-x|.

-2|5-x|<-6

 

Divide each term in the inequality by -2.  When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

-(2|5-x|)/(-2)>-(6)/(-2)

 

Simplify the left-hand side of the inequality by canceling the common factors.

|5-x|>-(6)/(-2)

 

Simplify the right-hand side of the inequality by simplifying each term.

|5-x|>3

 

Remove the absolute value term.  This creates a \ on the right-hand side of the equation because |x|=\x.

5-x>\(3)

 

Set up the + portion of the \ solution.

5-x>3

 

Move all terms not containing x to the right-hand side of the inequality.

-x>-2

 

Multiply -x by -1 to get x.

x<-2*-1

 

Divide each term in the inequality by -1.

x<2

 

Set up the – portion of the \ solution.  When solving the – portion of an inequality, flip the direction of the inequality sign.

5-x<-(3)

 

Multiply -1 by the 3 inside the parentheses.

-x+5<-3

 

Since 5 does not contain the variable to solve for, move it to the right-hand side of the inequality by subtracting 5 from both sides.

-x<-5-3

 

Subtract 3 from -5 to get -8.

-x<-8

 

Multiply each term in the inequality by -1.  When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

-x*-1>-8*-1

 

Multiply -x by -1 to get x.

x>-8*-1

 

Multiply -8 by -1 to get 8.

x>8

 

The solution to the inequality includes both the positive and negative versions of the absolute value.

x<2 or x>8

 

 

2.

1.  {(5,6),(5,7),(6,6),(6,7)}

(5,6),(5,7),(6,6),(6,7)

 

The domain of the set is the all values of x.  The range is the set of all values of y.

Domain: 5,6_Range: 6,7

 

g(x)=x^2-10x-3

f(x)=x^(2)-10x-3

 

The domain of the rational expression is all real numbers except where the expression is undefined.  In this case, there is no real number that makes the expression undefined.

All real numbers

 

f(x)=x^(2)-10x-3

 

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

x^(2)-10x-3=f(x)

 

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.

x^(2)-10x-f(x)-3=0

 

Use the quadratic formula to find the solutions.  In this case, the values are a=1, b=-10, and c=-1f(x)-3.

x=(-b\√(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

 

Use the standard form of the equation to find a, b, and c for this quadratic.

a=1, b=-10, and c=-1f(x)-3

 

Substitute in the values of a=1, b=-10, and c=-1f(x)-3.

x=(-(-10)\√((-10)^(2)-4(1)(-1f(x)-3)))/(2(1))

 

Multiply -1 by each term inside the parentheses.

x=(10\√((-10)^(2)-4(1)(-1f(x)-3)))/(2(1))

 

Simplify the section inside the radical.

x=(10\2√(f(x)+28))/(2(1))

 

Simplify the denominator of the quadratic formula.

x=(10\2√(f(x)+28))/(2)

 

First, solve the + portion of \.

x=(10+2√(f(x)+28))/(2)

 

Simplify the expression to solve for the + portion of the \.

x=5+√(f(x)+28)

 

Next, solve the – portion of \.

x=(10-2√(f(x)+28))/(2)

 

Simplify the expression to solve for the – portion of the \.

x=5-√(f(x)+28)

 

The final answer is the combination of both solutions.

x=5+√(f(x)+28),5-√(f(x)+28)

 

The domain of an expression is all real numbers except for the regions where the expression is undefined.  This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0.  All of these are undefined and therefore are not part of the domain.

(f(x)+28)<0

 

Solve the equation to find where the original expression is undefined.

f(x)<-28

 

The domain of the rational expression is all real numbers except where the expression is undefined.

f(x)>=-28_[-28,<Z>I<z>)

 

The domain of the inverse of f(x)=x^(2)-10x-3 is equal to the range of f(f(x))=5+√(f(x)+28).

Range: f(x)>=-28_[-28,<Z>I<z>)

 

 

 

 

 

 

 

3.

The vertical line test fails as the line creates two points on the circle,

So it Is not a function

4.

F(x)=x^3-x

f(x)=x^(3)-x

 

To check if a function is even, substitute in -x for x and see if the resulting function is the same as the original.  In other words, f(-x)=f(x).

f(-x)=(-x)^(3)-(-x)

 

Cubing an expression is the same as multiplying the expression by itself 3 times.

f(-x)=((-x)(-x)(-x))-(-x)

 

Multiply -x by -x to get x^(2).

f(-x)=((x^(2))(-x))-(-x)

 

Multiply x^(2) by -x to get -x^(3).

f(-x)=((-x^(3)))-(-x)

 

Remove the parentheses around the expression -x^(3).

f(-x)=(-x^(3))-(-x)

 

Multiply -1 by each term inside the parentheses.

f(-x)=-x^(3)+x

 

The function is not even because the resulting function (after substituting in x) is not the same as the original.

f(x)=x^(3)-x is not even.

 

To check if the function is odd, substitute in -x for x and check if the resulting function is the opposite of original function.  In other words, determine if f(-x)=-f(x).

f(-x)=(-x)^(3)-(-x)

 

Cubing an expression is the same as multiplying the expression by itself 3 times.

f(-x)=((-x)(-x)(-x))-(-x)

 

Multiply -x by -x to get x^(2).

f(-x)=((x^(2))(-x))-(-x)

 

Multiply x^(2) by -x to get -x^(3).

f(-x)=((-x^(3)))-(-x)

 

Remove the parentheses around the expression -x^(3).

f(-x)=(-x^(3))-(-x)

 

Multiply -1 by each term inside the parentheses.

f(-x)=-x^(3)+x

 

The function is odd because f(-x) is the opposite of f(x).  In other words, f(-x)=-f(x).

The function is odd.

 

So in the graph we see

 

F(1)  = 1 – 1 = 0

 

F(-1) = -1 – 1 = -2

 

So , these are not equal and don’t create a symmetry

 

 

U5

1.

(1,3),(2,4)

 

Slope is equal to the change in y over the change in x, or ‘rise over run’.

m=(change  in  y)/(change  in  x)

 

The change in x is equal to the difference in x-coordinates (also called run), and the change in y is equal to the difference in y-coordinates (also called rise).

m=(y2-y1)/(x2-x1)

 

Substitute in the values of x and y into the equation to find the slope.

m=(4-(3))/(2-(1))

 

Multiply -1 by the 1 inside the parentheses.

m=(4-(3))/(2-1)

 

Subtract 1 from 2 to get 1.

m=(4-(3))/(1)

 

Dividing any expression by 1 does not change the value of the expression.

m=(4-(3))

 

Multiply -1 by the 3 inside the parentheses.

m=4-3

 

Subtract 3 from 4 to get 1.

m=1

 

2.

y- (-3) = 2 * (x-0)

y = 2x + 3

 

3.

Since the lines are perpendicular

So slope = 1/ – 1 * – (-2) = ½

And the line passes to (1,2)

So the straight line is

(y-2)=1/2*(x-1)

y – 2 = ½ x – ½

y = ½ x + 3/2

 

slope intercept point

y= mx+ b

= 1/2x + b

It is passing through

(1,2)

 

So

2= ½+b

B=3/2

 

4.

Y=-5x+4

Parralel to line,

So slope is similar

So the straight passing through the equation

Y +7 = -5 (x+2 )

Y=-5X- 17

 

Slope intercept

Y= mx+ b

So putting the values

-7 = -5*-2 +b

-7 = b+10

B=17

SO.

STRAIGHT LINE

y= -5x -17

 

 

 

 

 

 

 

5.

F(x) =|x|

And as it is changed -|x| it is reflected in x axis and y  co-ordinates changes

G(X) =-|X+4|+2

The graph translated to |x+4| i.e  4  units to the left

And  it translates 2 units upward

 

 

 

 

 

 

 

 

For this graph

 

We can see increasing function is

From (-4,1)

 Source

 

 

 

 

Google Technologies

Which five Google technologies would you like to research for your Final Case Studies?

 

Google has paved the way for innovation by creating new web based and creative technology benefiting people all around the Globe. The Five most famous Technologies that I would love to research are as follows.

  1. Android Os.
  2. Google Glass
  3. Google Driverless
  4. Google Maps
  5. Google Street View

These are most amazing technologies Google have implement that has changed the world to a new place, after its amazingly successful Search Engine and web services.

 

 

  1. Share one interesting finding about each technology that you have chosen with the class.

 

The most interesting findings about each technology is that , The are most unique and different from each other  in respect .

 

1. Android OS : Google has made Android the coolest mobile OS in the world . With its advent of App store, Google play , Android has been one OS with its reach to anything in the world from Plane fares to Stock Prices in Entertainment, Technology , Education . The Google Play store has made Android a platform to run an App and excel in multitasking on Mobile and Smartphone devices.

 

  1. Google Glass: the most interesting thing about Google Glass is that its display information we need in front of our eyes  and with the tone of our voice it automates. This state of the art technology is one interesting finding of Google.

 

  1. Google Driverless Car: The Google Driverless car is one of its innovative projects, where the car is run by automation and a Camera to detect traffic and GPS system. The amazing feature is the future of automotive excellence.

 

  1. Google Maps : The Google Maps is most popular technology where its depicts any location along the globe and gives direction from any nearby  location by itself. It’s an all-important technology available in smartphones and mobiles and Tablets.

 

  1. Google Street View: The Google Street View is one of its Kind 3d Street View, that will enable you to visit places which you have never been to and the surroundings you have not seen on the Net. The

 

 

 

 

 

3. Which of these technologies do you think has the greatest potential to improve our society in the future? Why?

 

All the Technologies of Google are amazing helpful to the society in terms of Knowledge, Education, and Information and economy. I feel the most innovative of them is the Google Glass which can be later on modified to form various enhancements and be useful to billons of people all around the Globe. The Google glass has its user-friendly way to accessing the whole Internet world by voice commands, thus it has to adapt with different languages of different countries to make its User interface and it can be use to billions of people in the future.

 

 

 

 

 

References

 

  1. http://www.businessinsider.com/watch-this-blind-man-take-a-trip-in-one-of-googles-self-driving-cars-2012-3

 

  1. http://www.google.com/intl/en/about/products/index.html
  2. http://research.google.com/
  3.  http://www.GoogleX.com/