MATHEMATICS HIGH SCHOOL

Answer question number 5

Answers

Answer 1

Answer:

Answer:

36 is answer

Step-by-step explanation:

A cube has 6 sides.

216÷6=36

so therefore one side is 36 cubic inches

Answer 2

Answer:

$\tt Step-by-step~explanation:$

$\tt Step~1:$

Let's solve for one side first. The equation to solve for one side with only the volume provided is V ^ 1/3.

$\tt 216^{(1)/(3)} \n216^{(1)/(3)} =6$

One side of the cube is 6, so that means all sides of the cube is equal to 6.

$\tt Step~2:$

To find the area of one side of the cube, we take one side and multiply it by itself, since the formula for solving for the area is length • width.

$\tt 6*6=36$

$\large\boxed{\tt Our~final~answer:Area~of~one~side=36~in.^2}$

To check our answer, we can multiply 36 by 6 and see if it equals to 216.

$\tt 36*6=216=correct$

Answer:

Option: d is the correct answer.

d. 15

Step-by-step explanation:

We are given a exponential function as:

$f(x)=(1)/(5)(15^x)$

We know that in general a exponential function is represented by:

$f(x)=ab^x$

where a is the initial amount and b represents a growth factor if it is strictly greater than 1 and a decay factor if 0<b<1

Hence, the value of the growth factor of the function is:

d. 15

The answer would have to be 15.

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

Answers

$(tan^2x)/(1+cot^2x)+(cot^2x)/(1+tan^2x)=sec^2x\ cosec^2x-3\n\nL=(tan^2x(1+tan^2x)+cot^2x(1+cot^2x))/((1+cot^2x)(1+tan^2x))=(tan^2x+tan^4x+cot^2x+cot^4x)/(1+tan^2x+cot^2x+tanxcotx)\n\n=(tan^2x+cot^2x+tan^4x+cot^4x)/(1+tan^2x+cot^2x+1)=(tan^2x+2+cot^2x+tan^4x-2+cot^4x)/(tan^2x+cot^2x+2)$

$=((tanx+cotx)^2+(tan^2x-cot^2x)^2)/((tanx+cotx)^2)=((tanx+cotx)^2)/((tanx+cotx)^2)+((tan^2x-cot^2x)^2)/((tanx+cotx)^2)\n\n=1+((tanx-cotx)^2(tanx+cotx)^2)/((tanx+cotx)^2)=1+(tanx-cotx)^2\n\n=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\n\n=\left((sinx)/(cosx)\right)^2+\left((cosx)/(sinx)\right)^2-1=(sin^2x)/(cos^2x)+(cos^2x)/(sin^2x)-1=(sin^4x+cos^4x)/(sin^2x\ cos^2x)-1$

$=((sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x)/(sin^2x\ cos^2x)-1\n\n=((sin^2x+cos^2x)^2-2sin^2x\ cos^2x)/(sin^2x\ cos^2x)-1=(1^2-2sin^2x\ cos^2x)/(sin^2x\ cos^2x)-1\n\n=(1)/(sin^2x\ cos^2x)-(2sin^2x\ cos^2x)/(sin^2x\ cos^2x)-1=(1)/(sin^2x)\cdot(1)/(cos^2x)-2-1\n\n=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

ΔCAR has coordinates C (2, 4), A (1, 1), and R (3, 0). A translation maps point C to C' (3, 2). Find the coordinates of A' and R' under this translation. A' (4, −2); R' (2, −1) A' (−2, 2); R' (2, −2) A' (2, −1); R' (4, −2) A' (−1, 0); R' (−2, 2)

Answers

For the answer to the question above asking to find the coordinates of A' and R' under this translation if ΔCAR has coordinates C (2, 4), A (1, 1), and R (3, 0). A translation maps point C to C' (3, 2).
The answer is to the question above is A' (2, -1); R' (4, -2)
I hope my answer helped you. Have a nice day!

Answer:

A' (2, -1); R' (4, -2)

Step-by-step explanation:

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