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3 years ago

6

What is the range of the function graphed below? PLEASE HELP ASAP

1 answer:

DanielleElmas [232]3 years ago

4 0

I think it's A..... Iiiiiiiiiiiii

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A photo is 4 inches by 6 inches. What is the area of the photo?

kolezko [41]

4 * 6 = 24
The answer is, the area of the whole photo is 24.
Hope this helped :)

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3 years ago

A frustum is formed when a plane parallel to a cone’s base cuts off the upper portion as shown. Which expression represents the

Bezzdna [24]

Think of the two pieces together as a single cone of height 19, and its volume.
Then think of only the upper piece as a cone, and find its volume.
The volume of the frustum is the total volume minus the volume of the upper piece.

The volume of a cone is:

$V = \dfrac{1}{3} \pi r^2 h$

Total volume:

$V_t = \dfrac{1}{3} \times \pi \times 7.5^2 \times 19$

Volume of upper cone:

$V_u = \dfrac{1}{3} \times \pi \times 3.5^2 \times 8$

Volume of frustum:

$V_f = \dfrac{1}{3}\pi(7.5^2)(19) - \dfrac{1}{3}\pi(3.5^2)(8)$

3 0

3 years ago

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PLEASE HELP!!! I ONLY HAVE A FEW MINUTES!!

TEA [102]

Answer:

y=0.25x-6

Step-by-step explanation:

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2 years ago

On a coordinate plane, rhombus W X Y Z is shown. Point W is at (7, 2), point X is at (5, negative 1), point Y is at (3, 2), and

Otrada [13]

Answer:

$P = 4\sqrt{13}$

Step-by-step explanation:

Given

$W = (7, 2)$

$X = (5, -1)$

$Y = (3, 2)$

$Z =(5, 5)$

Required

The perimeter

To do this, we first calculate the side lengths using distance formula

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2$

So, we have:

$WX = \sqrt{(5- 7)^2 + (-1 - 2)^2$

$WX = \sqrt{13}$

$XY = \sqrt{(3-5)^2 + (2--1)^2}$

$XY = \sqrt{13}$

$YZ = \sqrt{(5-3)^2 + (5-2)^2}$

$YZ = \sqrt{13}$

$ZW = \sqrt{(7 - 5)^2 + (2 - 5)^2}$

$ZW = \sqrt{13}$

The perimeter is:

$P = WX + XY + YZ + ZW$

$P = \sqrt{13}+\sqrt{13}+\sqrt{13}+\sqrt{13}$

$P = 4\sqrt{13}$

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3 years ago

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15,266,783.93 + 6,332,682.27442 - 92.833 + 20/100 divided by 42

borishaifa [10]

Answer: 21599373.37618

Step-by-step explanation:

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2 years ago