MATHEMATICS HIGH SCHOOL

What is the factored form of x^2-36z^2?

Answers

Answer 1
Answer: x^2-36z^2

(x-6z)(x+6z)

because...

x*x=x^2

x*6z=6xz \n -6z*x=-6xz \n 6xz-6xz=0

-6z*6z=-36z^2

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What is the contrapositive of the statement? All squares are rectangles. A.If a figure is a square, then it is a rectangle.
B.If a figure is a rectangle, then it is a square.
C.If a figure is not a square, then it is not a rectangle.
D.If a figure is not a rectangle, then it is not a square.

Answers

D.) If a figure is not a rectangle, then it is not a square
Contrapositive is when it switches the two sides up and makes it negative as well.

Answer:

D

Step-by-step explanation:

What is the slope of the line 4x + 6y = 12

Answers

I believe its going to be (3,2) because you take 12 and divide it by 4 which gives you your x value of 3 then you take 12 and divide it by 6 which gives you your y value of 2
4x+6y=12
-4x        -4x
6y=-4x+12
-6          -6

Answer: Y= -4x+12

Length -12p + 4 width -7p + 10. Area ?

Answers

Answer:
84p^2-148p+40

A=LxW
=(-12p+4)(-7p+10)
=84p^2-148p+40

Suppose the radius of a circle is 8. what is its circumference

Answers

16 because the circumference is 2x the amount of the radius :) please give 5 stars and a thankyou button :)
First, here is the formula for finding circumference using the RADIUS:
C (circumference) = 2 x pi (3.14) x r (radius)
using the DIAMETER:
C = pi (3.14) x d (diameter)
------------
Using the above formula, just plug in the numbers!!
C = 2 x (3.14) x (8)
C = 50.24
50.24 is your answer

if you wanted to use diameter, you should know that diameter is just the radius times 2!

Which number is a solution of the inequality? 10.6 < ba. 14
b. –9
c. 7
d. –18

Answers

The only answer choice that is greater than 10.6 is letter A, 14

I need help combining functions!

Answers

\bf \begin{cases}f(x)=\cfrac{2}{x}\n\ng(x)=\cfrac{x}{x+2}\end{cases}\qquad g[\quad f(x)\quad ]=\cfrac{[f(x)]}{[f(x)]+2}

\bf (g\circ f)(x)\iff g[\quad f(x)\quad ]=\cfrac{\boxed{(2)/(x)}}{\boxed{(2)/(x)}+2}\n\n\ng[\quad f(x)\quad ]=\cfrac{(2)/(x)}{(2x+2)/(x)}\implies g[\quad f(x)\quad ]=\cfrac{2}{x}\cdot \cfrac{x}{2x+2}\n\n\ng[\quad f(x)\quad ]=\cfrac{2x}{2x^2+2x}\implies g[\quad f(x)\quad ]=\cfrac{2x}{2x(x+1)}\n\n\ng[\quad f(x)\quad ]=\cfrac{1}{x+1}
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