Complete the function table and write the rule.
Complete the function table and write the rule. - 1

Answers

Answer 1
Answer: Input  Output
32       20
14        2
?         -6
-2          -14
-10       ?
OK,  With the ones I have gotten full answers on,  the pattern is y=x-12
So,  ? one is 6 because you ADD 12.
? 2 on the y side is -22 because you subtract 12!

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What is 6:12 in simplest form

Answers

6/12 in simplest form is 1/2

In simplest form, the ratio 6:12 is equivalent to 1:2.

To simplify the ratio 6:12, we need to find the greatest common divisor (GCD) of the two numbers and then divide both numbers by the GCD.

Step 1: Find the GCD of 6 and 12.

The factors of 6 are 1, 2, 3, and 6.

The factors of 12 are 1, 2, 3, 4, 6, and 12.

The greatest common divisor (GCD) of 6 and 12 is 6.

Step 2: Divide both numbers in the ratio by the GCD (6).

6 ÷ 6 = 1

12 ÷ 6 = 2

Step 3: Write the simplified ratio as 1:2.

In simplest form, the ratio 6:12 is equivalent to 1:2. This means that both numbers in the ratio have been divided by their greatest common divisor to give the simplest representation. The simplified ratio 1:2 indicates that there is one unit of the first quantity for every two units of the second quantity.

To know more about ratio:

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3(4+6)55 x 2/3=


What is the answer?

Answers

Answer:

1100

Step-by-step explanation:

hope it helped

A triangle has one angle that measures 35°. which values could be the measures of the other two angles?

Answers

your answer to it is 90

Given the position function, s of t equals t cubed divided by 3 minus 12 times t squared divided by 2 plus 36 times t , between t = 0 and t = 15, where s is given in feet and t is measured in seconds, find the interval in seconds where the particle is moving to the right. (5 points)0 < t < 6


5 < t < 15


t > 15


The particle always moves to the right.

Answers

Answer:

Option D is correct.

as the particle always moves to the right

Step-by-step explanation:

Given the function:  s(t) = (t^3)/(3) - (12t^2)/(2) + 36t , 0<t<15;

where

s(t) represent the distance in feet.

t represents the time in second.

Since, when the particle is moving to the right,

⇒ s(t) is increasing.

so, v(t) =(ds)/(dt) > 0

Find the derivative of s(t) with respect to t;

Use derivative formula:

(dx^n)/(dx) = nx^(n-1)

(ds)/(dt) = (3t^2)/(3)-(24t)/(2)+36

Simplify:

(ds)/(dt) = t^2-12t+36

As (ds)/(dt) > 0

t^2-12t+36 > 0

(t-6)^2 >0               [(a-b)^2 = a^2-2ab+b^2 ]

⇒ this is always true because square of any number is always positive

Therefore, it means that the particle always moves to the right.

What is the surface area of the prism? A. 138 yd2 B. 192 yd2 C. 264 yd2 D. 272 yd2

Answers

In order to find the area of the triangular prism, we need to find the area of each of the figures that make it up. It might help if you draw the triangles and rectangles that make up the prism individually.

Let's start with the two triangles at either end of the prism.

A = 1/2bh   Area of a triangle
A = 1/2(6)(8)   Substitute
A = 24   Multiply / Divide

We now know that the area of the triangle at the end of the prism is 24 yd². However, since there are two of these triangles, we'll need to multiply that by 2. So, in total, the area of the two triangles is 48 yd².

There are three more rectangles that we need to find the area for: the base of the prism, the vertical edge of the prism, and the slanted edge of the prism. 

A = lw   Area of a rectangle
A = (9)(10)   Area of the slanted edge
A = 90 yd²   Multiply

A = (9)(6)   Area of the base
A = 54 yd²   Multiply

A = (8)(9)   Area of the vertical edge
A = 72 yd²

We now have the area for both triangles and each rectangle. Now, all we have to do is add these together to get the total surface area!

72 + 54 + 90 + 48 = 264 yd²
Surface Area = 264 yd²

The answer is 264 yd², or answer choice C.

My answers are wrong plz help

Answers

I think a would be 115 degrees. B would be 71 degrees. I think the second question would be c.

Your first answer is wrong.

The area should be:

a= 65

b= 109

The second one is probably C.)