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tangare [24]
4 years ago
7

A wheel of a gun wagon had a radius of 18 inches and was rotating at 40 rpm what was the speed of the wagon in miles/hours

Mathematics
1 answer:
matrenka [14]4 years ago
5 0
14 miles per hour  USE THE FORMULA! the answer will be yours
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Write about how to use the dollar sign and decimal point to show the total value of 5 quarters
aivan3 [116]
So 1 quater is 25 cents each. 5 quaters would be 125 cents. We know that 4 quaters equal $1.00 So we now have 1 quater left, you can add 25 cents to one dollar and get $1.25
So we started off with 125 cents so we added in a decimal point and added in a dollar sign. Hope this helps.
6 0
3 years ago
Simplify the following 7 times the square root of 54 - 2 times the square root of 24
EleoNora [17]
The first step for solving this problem is to simplify the radical
7\sqrt{54 - 2} x 2√6
Calculate the product
14\sqrt{(54 - 2) X 6}
Subtract the numbers
14\sqrt{56 x 6}
Simplify the radical
14 x 2\sqrt{13 x 6}
Multiply the numbers
14 x 2√78
Finally,, calculate the product for your final answer
28√78
Let me know if you have any further questions
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4 0
4 years ago
HELPPPPPPPPPPP!!!!!!!!!!!!!
Molodets [167]
Choice A is false. There are 15% of freshman in her BIO class, but there are 35% of her total students in it.
3 0
4 years ago
If a substance decays at a rate of 25% every 10 years, how long will it take 512 grams of the substance to decay to 121.5 grams
Juliette [100K]

Answer:

It will take 50 years to decay from 512 grams to 121.5 grams.

Step-by-step explanation:

The decay formula :

N=N_0e^{-\lambda t}

where

N= amount of substance after t time

N₀= initial of substance

t= time.

A substance decays at a rate 25% every 10 years.

So, remaining amount of the substance is = (100%-25%)= 75%

\frac{N}{N_0}=\frac{75\%}{100\%}=\frac{75}{100}=\frac34, t= 10

N=N_0e^{-\lambda t}

\Rightarrow \frac {N}{N_0}=e^{-\lambda t}

\Rightarrow \frac34 =e^{-\lambda .10}

Taking ln both sides

\Rightarrow ln|\frac34| =ln|e^{-\lambda .10}|

\Rightarrow ln|\frac34|=-10\lambda

\Rightarrow \lambda=\frac{ ln|\frac34|}{-10}

Now , N₀= 512 grams, N= 121.5 grams, t=?

N=N_0e^{-\lambda t}

\therefore 121.5=512e^{-\frac{ln|\frac34|}{-10}.t}

\Rightarrow 121.5=512e^{\frac{ln|\frac34|}{10}.t}

\Rightarrow \frac{121.5}{512}=e^{\frac{ln|\frac34|}{10}.t}

Taking ln both sides

\Rightarrow ln|\frac{121.5}{512}|=ln|e^{\frac{ln|\frac34|}{10}.t}|

\Rightarrow ln|\frac{121.5}{512}|={\frac{ln|\frac34|}{10}.t}

\Rightarrow t=\frac{ln|\frac{121.5}{512}|}{\frac{ln|\frac34|}{10}}

\Rightarrow t=\frac{10.ln|\frac{121.5}{512}|}{{ln|\frac34|}}

⇒t=50 years

It will take 50 years to decay from 512 grams to 121.5 grams.

8 0
4 years ago
What is the product of one half and two thirds?
Grace [21]

Answer:

Step-by-step explanation:

0.33333333333

7 0
3 years ago
Read 2 more answers
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