Choose the graph 12-4×>8

HELP! The diagonal length of a kite is multiplied by 8. Which of the following describes the effect of this change on the area?

jeka57 [31]

Answer:

The area is multiplied by 8.

Step-by-step explanation:

I am pretty sure I multiplied the two diagonals that I randomly selected then multiplied the longest diagonal by eight multiples them together again then divided the bigger area by the smaller area, however, but you might want to double-check the work but the answer is right

6 0

3 years ago

A volleyball of radius 13 inches fits exactly in a cylinder with the same radius. inside the container there is empty space not

asambeis [7]

Volume of Sphere is V=(4/3)*PI*r^3

The radius of the volleyball is 13 inches

So V=(4/3) * 3.14 * 13^3 = 9198 cubic inches

Volume of a cylinder is V=pi * r^2 *h

Since they say the ball fits perfectly

The radius is 13 and the height is 26 ( the diameter of the ball)

So V=3.14 * 13^2 * 26 = 13,797 cubic inches

13797 – 9198 = 4,599 cubic inches of empty space

4 0

3 years ago

Read 2 more answers

A soup can in the shape of a right circular cylinder is to be made from two materials. The material for the side of the can cost

Advocard [28]

Answer:

Radius = 1.12 inches and Height = 4.06 inches

Step-by-step explanation:

A soup can is in the shape of a right circular cylinder.

Let the radius of the can is 'r' and height of the can is 'h'.

It has been given that the can is made up of two materials.

Material used for side of the can costs $0.015 and material used for the lids costs $0.027.

Surface area of the can is represented by

S = 2πr² + 2πrh ( surface area of the lids + surface are of the curved surface)

Now the function that represents the cost to construct the can will be

C = 2πr²(0.027) + 2πrh(0.015)

C = 0.054πr² + 0.03πrh ---------(1)

Volume of the can = Volume of a cylinder = πr²h

16 = πr²h

$h=\frac{16}{\pi r^{2}}$ -------(2)

Now we place the value of h in the equation (1) from equation (2)

$C=0.054\pi r^{2}+0.03\pi r(\frac{16}{\pi r^{2}})$

$C=0.054\pi r^{2}+0.03(\frac{16}{r})$

$C=0.054\pi r^{2}+(\frac{0.48}{r})$

Now we will take the derivative of the cost C with respect to r to get the value of r to get the value to construct the can.

$C'=0.108\pi r-(\frac{0.48}{r^{2} })$

Now for C' = 0

$0.108\pi r-(\frac{0.48}{r^{2} })=0$

$0.108\pi r=(\frac{0.48}{r^{2} })$

$r^{3}=\frac{0.48}{0.108\pi }$

r³ = 1.415

r = 1.12 inch

and h = $\frac{16}{\pi (1.12)^{2}}$

h = 4.06 inches

Let's check the whether the cost is minimum or maximum.

We take the second derivative of the function.

$C"=0.108+\frac{0.48}{r^{3}}$ which is positive which represents that for r = 1.12 inch cost to construct the can will be minimum.

Therefore, to minimize the cost of the can dimensions of the can should be

Radius = 1.12 inches and Height = 4.06 inches

5 0

3 years ago

Which function is increasing and has a vertical asymptote at x = 5?

marusya05 [52]

C. f(x)= ln(x-5) hope that helps

3 0

3 years ago

Which point is located at (-3,6) A) A B) B C) C D) D

Andrej [43]

Answer is A

Step-by-step explanation:

Go over to the left 3 times then up 6

3 0

3 years ago