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3 years ago

The circle has a circumference of 453.26 what is the radius

Mathematics

1 answer:

g100num [7]3 years ago

3 0

Answer:

72.138 ~ 72.1 as the radius of the circle.

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it took 8 hours for karl and lou to finish a project for science class together. if karl had worked alone on the project it woul

kkurt [141]

It would take 10 hours just like it did for Karl .

4 0

3 years ago

Please help me with this

horrorfan [7]

Answer:

$\dfrac{4000\pi}{3}$ ft³

Step-by-step explanation:

First, let's figure out how to get the volume of a sphere from its surface area. If r is the radius of our sphere, then

The formula for a sphere's surface area is $A = 4\pi r^2$

The formula for a sphere's volume is $V=\frac{4}{3}\pi r^3$

So to get from area to volume, we have to divide the area by 3 and then multiply it by r. Mathematically:

$V=\frac{A}{3}r$

Before we solve for V though, we need to find the radius of our sphere. Thankfully, we're given the surface area - $400\pi$ ft² - so we can use the area formula to find that radius:

$A=4\pi r^2=400\pi\\r^2=100\\r=10$

And now that we have our radius, we can put it into our volume formula to find

$V=\frac{A}{3} r=\frac{400\pi}{3}(10)=\frac{4000\pi}{3}$ ft³

4 0

3 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately 2020.022

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$