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3 years ago

I got the answer 0.007853982 inches sq, but it says it’s not correct, what am I doing wrong?

Mathematics

1 answer:

sesenic [268]3 years ago

7 0

Answer:

Area of pupil = 0.007855 square inch.

Step-by-step explanation:

Given the following data;

Diameter = 0.1 inch

To find the area of the pupil (circle);

Area of circle = π(D/2)²

Area of pupil = 3.142*(0.1/2)²

Area of pupil = 3.142*(0.05)²

Area of pupil = 3.142*0.0025

Area of pupil = 0.007855 square inch.

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Which of these describes an angle?

stealth61 [152]

A part of a line that connects two endpoints I believe

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2 years ago

Please help me What does m

Alja [10]

9514 1404 393

Answer:

∠N = 10°

Step-by-step explanation:

Assuming the marked point on segment LN is supposed to be the center of the circle, arc NML is 180°, so arc ML is 180° -160° = 20°.

Arc ML is intercepted by inscribed angle LNM, so the measure of that angle is half the arc measure:

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2 years ago

PLEASE HELP ASAP!!!

Elan Coil [88]

Answer:

The number is 5

Step-by-step explanation:

6+3n = 5n-4

Subtract 3n from each side

6+3n-3n = 5n-3n-4

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6 0

3 years ago

If a phone cost $224 what would it cost if there was a 7% sales tax.

Annette [7]

Answer:

$239.68

Step-by-step explanation:

Given

Cost = $224
Sales Tax = 7%

Solving

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5 0

2 years ago

Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circu

Sunny_sXe [5.5K]

Answer:

0.0221 feet per minute.

Step-by-step explanation:

$\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h$

If the Base Diameter = Height of the Cone

The radius of the Cone = h/2

Therefore,

$\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}$

Rate of Change of the Volume, $\dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}$

Since gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. Therefore, the Volume of the cone is increasing at a rate of 10 cubic feet per minute.

$\dfrac{dV}{dt}=10$ ft^3/min$

We want to determine how fast is the height of the pile is increasing when the pile is 24 feet high.

We have:

$\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=10\\\\$When h=24$ feet$\\\dfrac{3\pi *24^2}{12}\dfrac{dh}{dt}=10\\144\pi \dfrac{dh}{dt}=10\\ \dfrac{dh}{dt}= \dfrac{10}{144\pi}\\ \dfrac{dh}{dt}=0.0221$ feet per minute$

When the pile is 24 feet high, the height of the pile is increasing at a rate of 0.0221 feet per minute.

4 0

3 years ago