MATHEMATICS HIGH SCHOOL

According to the Rational Root Theorem, the following are potential fox) 2x2 +2x 24.roots of -4, -3, 2, 3, 4Which are actual roots of f(x)?O 4 and 34, 2, and 3O 3 and 43, 2, and 4

Answers

Answer 1

Answer:

Correct Answer:First Option

Explanation:

There are two ways to find the actual roots:

a) Either solve the given quadratic equation to find the actual roots

b) Or substitute the value of Possible Rational Roots one by one to find out which satisfies the given equation.

Method a is more convenient and less time consuming, so I'll be solving the given equation by factorization to find its actual roots. To find the actual roots set the given equation equal to zero and solve for x as given below:

$2x^(2) +2x-24=0\n \n 2(x^(2) +x-12)=0\n \n x^(2) +x-12=0\n \n x^(2) +4x-3x-12=0\n \n x(x+4)-3(x+4)=0\n \n (x-3)(x+4)=0\n \n x-3=0, x=3\n \n or\n\nx+4=0, x=-4$

This means the actual roots of the given equation are 3 and -4. So first option gives the correct answer.

Answer 2

Answer: The correct answer for the question that is being presented above is this one: "-4 and 3." According to the Rational Root Theorem, the following are potential f(x) 2x2 +2x 24. The actual roots of the quadratic equation based on the Rational Root Theorem are -4 and 3.

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HELP PLEASE!! What are the domain and range of each relation? Drag the answer into the box to match each relation

Answers

Answer: (1) The domain and range of figure is shown in option 3. (2) The domain and range of the coordinate pairs is shown in option 2.

Explanation:

(1)

The values of x for which the function is defined are the domain of that function and range is the resultant values of the function at different value of x.

From The given figure it is noticed that the coordinates of the points are,

$(-3,1),(-3,-4),(-1,1),(1,-1),(4,4)$

A coordinate pair is defined as (x,y).The values of x are -3,-3,-1,1,4. All sets contains only distinct elements so repetition of elements is not allowed.

Domain = {-3,-1,1,4}

The values of y are 1,-4,1,-1,4.

Range = {-4,-1,1,4]

So the correct option is third.

(2)

The given coordinates are,

$(-4,3),(-4,4),(-3,1),(1,1)$

The values of x are -4,-4,-3,1.

Domain = {-4,-3,1}

The values of y are 3,4,1,1.

Range = {1,3,4}

Therefore the correct option is (2).

An item is purchased for a wholesale price of $48 and will be sold at a 55 percent markup. Which expression should be used to find the amount of the markup?

Answers

Answer:

The answer is 48(0.55) x 48

Answer:

The answer is 48(0.55) x 48

Step-by-step explanation:

wholesale $48 and sold at 55% which is the same as 0.55 then multiply wholesale by percent

Ex 3.7
12. find the area between the curve y=x³-2 and the y-axis between y= -1 and y=25

Answers

$y=x^3-2\nx^3=y+2\nx=\sqrt[3]{y+2}\n\n\int \limits_(-1)^(25)\sqrt[3]{y+2}\, dy=\n\int \limits_(-1)^(25)(y+2)^{\tfrac{1}{3}}\, dy=\n\left[\frac{(y+2)^{\tfrac{4}{3}}}{(4)/(3)} \right]_(-1)^(25)=\n$
$\left[(3)/(4)(y+2)^{\tfrac{4}{3}} \right]_(-1)^(25)=\n\left[(3)/(4)(y+2)\sqrt[3]{y+2} \right]_(-1)^(25)=\n(3)/(4)(25+2)\sqrt[3]{25+2}-\left((3)/(4)(-1+2)\sqrt[3]{-1+2}\right)=\n(3)/(4)\cdot27\sqrt[3]{27}-\left((3)/(4)\sqrt[3]{1}\right)=\n(3)/(4)\cdot27\cdot3-(3)/(4)=\n(3)/(4)(81-1)=\n(3)/(4)\cdot 80=\n3\cdot20=\n60$

Yeah, you'd have to use the inverse function to produce this result.

Let's get the inverse function first:

$y={ x }^( 3 )-2\n \n { x }^( 3 )=y+2\n \n x=\sqrt [ 3 ]{ y+2 }$

$\n \n \therefore \quad { f }^( -1 )\left( x \right) =\sqrt [ 3 ]{ x+2 }$

Now, we can solve the problem using:

$\int _( -1 )^( 25 ){ \sqrt [ 3 ]{ x+2 } } dx$

But to solve the problem more easily we make u=x+2, therefore du/dx=1, therefore du=dx.

When x=25, u=27.

When x=-1, u=1.

Now:

$\int _( 1 )^( 27 ){ { u }^{ \frac { 1 }{ 3 } } } du\n \n ={ \left[ \frac { 3 }{ 4 } { u }^{ \frac { 4 }{ 3 } } \right] }_( 1 )^( 27 )$

$\n \n =\frac { 3 }{ 4 } \cdot { 27 }^{ \frac { 4 }{ 3 } }-\frac { 3 }{ 4 } \cdot { 1 }^{ \frac { 4 }{ 3 } }\n \n =\frac { 3 }{ 4 } { \left( { 3 }^( 3 ) \right) }^{ \frac { 4 }{ 3 } }-\frac { 3 }{ 4 }$

$\n \n =\frac { 3 }{ 4 } \cdot { 3 }^( 4 )-\frac { 3 }{ 4 } \n \n =\frac { 3 }{ 4 } \left( { 3 }^( 4 )-1 \right)$

$\n \n =\frac { 3 }{ 4 } \cdot 80\n \n =60$

Answer: 60 units squared.

−15x+4x+2−x=____x+ 3