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4 years ago

What is the arc length of the semi circle. hurry!!

Mathematics

1 answer:

svp [43]4 years ago

8 0

Answer:

Arc length $= \pi \times radius= \pi \times 6$

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From 5x^2-6x+8, subtract 3x^2-2x-4?

victus00 [196]

$(5x^2-6x+8)-(3x^2-2x-4)=\\5x^2-6x+8-3x^2+2x+4=2x^2-4x+12$

8 0

4 years ago

HELP ON #2 PLEASEEE!!!!!!!!!!!!!!!!!!!!!

garik1379 [7]

Answer:

ok so i think it is H

Step-by-step explanation:

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3 years ago

Read 2 more answers

Please answer these questions in the image ( click image to see fully )

drek231 [11]

9) x= -7

10) x= 10

11) x= 5

12) x= 6

13) x= 4

14) x= -9

15) x= 3

16) x= 9

sorry it took so long! hope this helps!! :)

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3 years ago

Mario's refrigerator is a mess. he discovers 3 large pizza boxes in there, all with some pizza in them. one has 1/2 of pizza, an

Fudgin [204]

Answer:

5 1/8 pizza

Step-by-step explanation:

3/4=6/8

1/2=4/8

7/8=7/8

6/8+4/8+7/8=17/8= 2 1/8 pizza

3+2 1/8= 5 1/8

8 0

3 years ago

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If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate at which the diameter decreases whe

mixas84 [53]

Answer:

The diameter decreases at the rate of 0.0796 cm/min when the diameter is 10 cm.

Step-by-step explanation:

A snowball is spherical, so it's area is given by the following formula:

$A = 4\pi r^{2}$

The radius is half the diameter so:

$A = 4\pi (\frac{d}{2})^{2}$

$A = 4\pi \frac{d^{2}}{4}$

$A = \pi d^{2}$

If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm.

We have to find $\frac{dd}{dt}$ when $\frac{dA}{dt} = -5, d = 10$

$A = \pi d^{2}$

Applying implicit differentiation:

We have to variables(A and d), so:

$\frac{dA}{dt} = 2\pi d \frac{dd}{dt}$

$-5 = 2\pi (10) \frac{dd}{dt}$

$\frac{dd}{dt} = -\frac{5}{20\pi}$

$\frac{dd}{dt} = -0.0796$

The diameter decreases at the rate of 0.0796 cm/min when the diameter is 10 cm.

6 0

4 years ago