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3 years ago

Which formula can you use to find the radius (r)r if you know that C=2πr?

1 answer:

3 0

First of all, it is 2r $\pi$ and it would be divide by 3.14 (pi) and then you should have the diameter. Then divide by 2 to find radius.

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Mason has a collection of 49¢ stamps, 20¢ stamps, and 34 stamps worth $23.55. He has 56 total 49¢ and 20+ stamps and the number

tatuchka [14]

Answer:

Mason has

16 20¢ stamps;
25 3¢ stamps;
40 49¢ stamps.

Step-by-step explanation:

Let x be the number of 20¢ stamps, then the number of 3¢ stamps is x + 9.

Mason has 56 total 49¢ and 20¢ stamps, then the number of 49¢ stamps is 56 - x.

Amounts:

in 49¢: $49(56-x)$ cents;
in 20¢: $20x$ cents;
in 3¢: $3(x+9)$ cents.

Total amount:

$49(56-x)+20x+3(x+9);$
$23.55.

Hence,

$49(56-x)+20x+3(x+9)=2,355\\ \\2,744-49x+20x+3x+27=2,355\\ \\-49x+3x+20x=2,355-2,744-27\\ \\-26x=-416\\ \\26x=416\\ \\x=16$

Mason has

16 20¢ stamps;
25 3¢ stamps;
40 49¢ stamps.

5 0

3 years ago

Only answer if you actually want to help me please

mamaluj [8]

the answer is circle idk if it's right but it could be I'm in high school I've been through algebra 1 now in geometry

5 0

3 years ago

Sam has 3/8 yard of twine to build model ship how much can he buy

Umnica [9.8K]

You never told us how much money he has

5 0

3 years ago

Help please, anything helps (answer which ever ones u can, you don't have to answer all of them just one is fine, but if u want

vivado [14]

Answer:

lol sry for AWFUL handwriting!!!! Hope this helps!! Have a great day : )

3 0

3 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

4 years ago