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)