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3 years ago

How to find the diameter of a circle with the area of 1134.1?

1 answer:

6 0

Radius is half of the diameter.
So because the area of a circle is A= pi • r^2, to find the radius first, we have to change the equation to solve for r. Now it is, r= square root of A/pi. We plug in 1134.1. (Look at pic)
The diameter is 38

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A student solved this problem and said the answer is quarts

kenny6666 [7]

B. no the answer isn't reasonable, it should be about 3 quarts.

3 0

3 years ago

Complete this activity. Mr. Home buys two quarts of ice cream, one carton of soda, and three sacks of cookies for a Sunday Schoo

makkiz [27]

Answer:

1300

Step-by-step explanation:

add 60x2+70+45x3

get 325

multaply times 4 because he gets this stuff then goes back and gets it 3 more times

get 1300

4 0

3 years ago

William doubled the amount of money he had after raking leaves. Now he has $24.50. How much money did he hav before he doubled i

ki77a [65]

Answer:

x=8.16

Step-by-step explanation:

x=the money before

2x=doubled

3x=24.50

x=8.16

7 0

3 years ago

Read 2 more answers

Solving Rational equations. LCD method. Show work. Image attached.

posledela

$(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12$

So in order to get all the fractions to have a common denominator, we need to multiply $\dfrac k{k-2}$ by $\dfrac{k-6}{k-6}$ , and $\dfrac1{k-6}$ by $\dfrac{k-2}{k-2}$ :

$\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}$

$\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}$

Now,

$\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}$

As long as $k\neq2$ and $k\neq6$ (which we can't have because otherwise $k^2-8k+12=0$ ), we can cancel $k^2-8k+12$ in the denominators on both sides:

$4=(k^2-6k)+(k-2)$

$4=k^2-5k-2$

$0=k^2-5k-6$

We can factorize the right side:

$0=(k-6)(k+1)$

which tells us that $k=6$ and $k=-1$ are solutions.

4 0

3 years ago

Iridium -192 has a half life of 74 days after 148 days, how many milligrams of 900 mg sample will remain?

Ad libitum [116K]

$\bf \textit{Amount for Exponential Decay using Half-Life}
\\\\
A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &900\\
t=\textit{elapsed time}\to &148\\
h=\textit{half-life}\to &74
\end{cases}
\\\\\\
A=900\left( \frac{1}{2} \right)^{\frac{148}{74}}\implies A=900\left( \frac{1}{2} \right)^2\implies A=225$

8 0

3 years ago