MATHEMATICS HIGH SCHOOL

What is the area of a rectangle with vertices at (2, 3), (7, 3), (7, 10), and (2, 10)?a. 44 units2
b. 30 units2
c. 24 units2
d. 35 units2

Answers

Answer 1
Answer: Given:
x     y
2     3
7     3
7   10
2   10

It is easier to graph these vertices to avoid confusion. 

Length: basis is the y-values : 10 - 3 = 7 units
Width: basis is the x-values: 7-2 = 5 units

Area = Length * Width
A = 7 units * 5 units
A = 35 units²    Choice D.

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Find the rational number halfway between 5/16 and 7/8

Answers

\left((5)/(16)+(7)/(8)\right):2=\left((5)/(16)+(14)/(16)\right):(2)/(1)=(5+14)/(16)\cdot(1)/(2)=(19)/(16)\cdot(1)/(2)=(19)/(32)

What is the factored form of the polynomial?

x2 − 12x + 27?

Answers

Answer:

The factors of the expression in the question x² − 12x + 27 = 0  are (x - 9)(x -3) .

Step-by-step explanation:

As given the expression in the question be as follow .

x² − 12x + 27 = 0

x² - 9x - 3x + 27 = 0

x (x - 9) -3 (x -9) = 0

(x - 9)(x -3) = 0

Therefore the factors of the expression in the question x² − 12x + 27 = 0  are (x - 9)(x -3) .

The factored form of the polynomial ( x² - 12x + 27 ) is ( x-3 )( x-9 ).

What is the factored form of the polynomial?

Given that;

  • x² - 12x + 27
  • Factored form = ?

First, we think of two numbers where their addition gives -12 and their multiplication gives 27.

-3 and -9 fits perfectly.

Hence we have;

x² - 12x + 27

( x-3 )( x-9 )

Therefore, the factored form of the polynomial ( x² - 12x + 27 ) is ( x-3 )( x-9 ).

Learn more about factorizations here: brainly.com/question/1863222

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The zeros of the function f(x)=3x^2-3x-6 are

Answers

3x^2-3x-6=0\n 3(x^2-x-2)=0\n x^2-x-2=0 x^2+x-2x-2=0\n x(x+1)-2(x+1)=0\n (x-2)(x+1)=0\n x=2 \vee x=-1

Two lines are parallel. The equation for the first line is y=-3x+5. Which of the following can't be the equation of the second line?A. y=-3x+2
B. x-3y=-3
C. 3x+y=-6
D. 6x+2y=2 Two lines are parallel. The equation for the first line is y=-3x+5. Which of the following can't be the equation of the second line?
A. y=-3x+2
B. x-3y=-3
C. 3x+y=-6
D. 6x+2y=2 @Mathematics

Answers

Parallel lines can be distinguished by looking at the value of their slopes. These equations should have the same value of the slope. We are given the equation,

y=-3x+5

From the choices, we see that A, C and D have the same slope with the equation. Therefore, the correct answer is B, it is the equation that can't be parallel with the given equation.

Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]. so far I found f'(x)=2cos(2x)+2 cos(2x)=-1

Answers

The solutions of the equation f\left( x \right) = 2x + \sin \left( {2x} \right) in the interval \left[ {0,2\pi } \right] are \boxed{\left( {(\pi )/(2),\pi } \right)} and \boxed{\left( {\frac{{3\pi }}{2},3\pi } \right)}.

Further explanation:

Given:

The function is f\left( x \right) = 2x + \sin \left( {2x} \right).

The first derivative is zero.

Explanation:

The given function is f\left( x \right) = 2x + \sin \left( {2x} \right).

Differentiate the function with respect to x.

\begin{aligned}f'\left( x \right) &= 2 + 2\cos \left( {2x} \right)\n&= 2\left( {1 + \cos 2x} \right)\n\end{aligned}

Substitute 0 for f'\left( x \right).

\begin{aligned}2\left( {1 + \cos 2x} \right) &= 0 \n1 + \cos 2x &= 0\n\cos 2x &= - 1\n2x &= {\cos ^( - 1)}\left( { - 1} \right)\n2x &= \frac{{\left( {2n - 1} \right)\pi }}{2} \n\end{aligned}

In the interval \left[ {0,2\pi } \right] the x-coordinates are \boxed{(\pi )/(2)}{\text{ and }}\boxed{\frac{{3\pi }}{2}}.

The solutions of the equation f\left( x \right) = 2x + \sin \left( {2x} \right) in the interval \left[ {0,2\pi } \right] are \boxed{\left( {(\pi )/(2),\pi } \right)} and \boxed{\left( {\frac{{3\pi }}{2},3\pi } \right)}.

Learn more:

  1. Learn more about inverse of the function brainly.com/question/1632445.
  2. Learn more about equation of circle brainly.com/question/1506955.
  3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Application of derivatives

Keywords: derivative, x – coordinates, interval, far, 2x, sin2x, coordinates, 0, 2pi, y-coordinate.
From there, you simply need algebra and a calculator that works in radians.

Take the inverse cos of both sides to get 2x = arccos(-1) 

Then divide both sides by 2 to get x = arccos(-1) / 2 

Put that into a calculator and you get π/2. But because your bounds are 0 to 2π, you have to add π your solution to get the solution on the other side of the unit circle, which would be (3π/2).

Now that you have the x value, put (π/2) and (3π/2) into f(x) to get the y coordinate. 

f(π/2) = 2(π/2) + sin(2(π/2) = π, which means this solution is just (π/2, π)
f(3π/2 = 2(3π/2) + sin(2(3π/2) = 3π, which means this solution is (3π/2, 3π)

Given the equation Square root of 8x plus 1 = 5, solve for x and identify if it is an extraneous solution.

Answers

It depends if you mean √(8x) +1=5 or √(8x+1) = 5 

√(8x) +1=5

Subtract 1 from both sides. 

√(8x) =4

Take the square root of both sides. 

8x = sqrt(4) 
8x = 2
x = 1/4.

Plugging it back in: 
sqrt(8*0.25) + 1 = 5 
sqrt(2) = 4

This is not true because the square root of 2 is about 1.41, which does not equal 4. 
So it is an extraneous solution.
__________________________________________
√(8x+1) = 5 
Square both sides. 
8x + 1 = 25 
8x = 24
x = 3

Not extraneous because equation works out when you plug x back in.
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