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zysi [14]
3 years ago
14

How much fencing woul you need to enclose a circular pond with diameter 12 feet use 3.14 for pi. Choices are in the image. PLEAS

E HELP​

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Answer:

37.68

Step-by-step explanation:

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If P=2L+2W, solve for W.
vodka [1.7K]

Answer:

W = (P-2L)/2

Step-by-step explanation:

Isolate W

P = 2L + 2W

subtract 2L from each side

P - 2L = 2W

divide each side by 2

(P-2L)/2 = W

3 0
3 years ago
Trapezoids A and B are similar. Find the values of x and y.
Wewaii [24]

Step-by-step explanation:

when 2 shapes are similar, it means that all their angles are the same compared with each other.

and all distances (incl. side lengths) if one shape correlate to the corresponding distances in the other shadow by the same ratio or factor.

we have one side (30 m in A and 50 m in B), where we have both measures.

we see that the ratio to "go" from A to B is 30/50 = 3/5.

that means a distance in A is the same distance in B multiplied by 3/5.

and that factor is the same also for every other side.

therefore,

x = 15 m × 3/5 = 3 m × 3 = 9 m

y × 3/5 = 18 m

3y = 90 m

y = 30 m

3 0
2 years ago
Tim worked 51 hours last week. He earns $15.20 per hour plus overtime, at the usual rate, for hours exceeding 40 hours what was
olchik [2.2K]
It is A 698.90 thanks
3 0
3 years ago
Please help and thank you
IrinaK [193]

Answer:

b and e

Step-by-step explanation:

5 0
3 years ago
4. Meagan invests $1,200 each year in an IRA for 12 years in an account that earned 5%
Nina [5.8K]
Part A)

\bf \qquad \qquad \textit{Future Value of an ordinary annuity}\\
\left. \qquad \qquad \right.(\textit{payments at the end of the period})
\\\\
A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]

\bf \qquad 
\begin{cases}
A=
\begin{array}{llll}
\textit{accumulated amount}\\
\end{array}
\begin{array}{llll}

\end{array}\\
pymnt=\textit{periodic payments}\to &1200\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &12
\end{cases}

\bf A=1200\left[ \cfrac{\left( 1+\frac{0.05}{1} \right)^{1\cdot  12}-1}{\frac{0.05}{1}} \right]\implies A\approx 19100.55

part B)

so, for the next 11 years, she didn't make any deposits on it and simple let it sit and collect interest, compounded annually at 5%.

\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\  
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$19100.55\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &11
\end{cases}
\\\\\\
A=19100.55\left(1+\frac{0.05}{1}\right)^{1\cdot  11}\implies A\approx 32668.42

part C)

well, for 12 years she deposited 1200 bucks, that means 12 * 1200, or 14,400.

now, here she is, 12+11, or 23 years later, and she's got 32,668.42 bucks?

all that came out of her pocket was 14,400, so 32,668.42 - 14,400, is how much she earned in interest.
6 0
4 years ago
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