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vova2212 [387]
3 years ago
15

Can some one help me with this math problem please?

Mathematics
1 answer:
-BARSIC- [3]3 years ago
8 0
Both the work and explanation is in the picture.

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Quadrilateral ABCD is similar to Quadrilateral EFGH. Diagonal AC has length 7 and diagonal EG has length 13. What is the scale f
olasank [31]

Answer:

\large \boxed{\text{A. }\dfrac{13}{7}\text{; B. }\dfrac{17}{14} \text{; C. 507 in}^{2}}

Step-by-step explanation:

A. Scale factor

When you dilate an object by a scale factor, you multiply its line lengths by the same number.

If EF/AB = 13/7, the scale factor is 13/7.

B. Length of EF

\begin{array}{rcl}\dfrac{EF}{AB} & = & \dfrac{13}{7}\\\\\dfrac{EF}{\frac{17}{26}} & = & \dfrac{13}{7}\\\\EF & = & \dfrac{13}{7}\times\dfrac{17}{26}\\\\ & = &\dfrac{1}{7}\times\dfrac{17}{2}\\\\ & = & \mathbf{\dfrac{17}{14}}\\\end{array}\\\text{The length of EF is $\large \boxed{\mathbf{ \dfrac{17}{14}}}$}

C. Area of EFGH

If the lengths in a shape are all multiplied by a scale factor, then the areas will be multiplied by the scale factor squared.

ABCD is dilated by a scale factor of 13/7, so its area is dilated by a scale factor of

\left(\dfrac{13}{7} \right)^{2} = \dfrac{169}{49}

The area of its dilated image EFGH is

\text{Area of EFGH} = \text{147 in}^{2} \times \dfrac{\text{169}}{\text{49}} = 3 \times 169\text{ in}^{2} = 507 \text{ in}^{2}\\\\\text{The area of EFGH is $\large \boxed{\textbf{507 in}^{\mathbf{2}}}$}

8 0
3 years ago
a map has a scale of 2 in : 6 mi if clayton and centerville are 10 in apart on the map then how far apart are the real cities
Ahat [919]
I"m pretty sure they would be 60 mi apart
5 0
3 years ago
11000 is compounded semiannually at a rate of 12% for 21 years. Find the total amount in the compound interest account
marishachu [46]

\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\dotfill &\$11000\\
r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semiannually, thus twice}
\end{array}\dotfill &2\\
t=years\dotfill &21
\end{cases}


\bf A=11000\left(1+\frac{0.12}{2}\right)^{2\cdot 21}\implies A=11000(1.06)^{42}\implies A\approx 127127.359416

5 0
4 years ago
Suppose 8 out of every 20 students are absent from school less than five days a year.How many students would be absent from scho
Paladinen [302]
...................333
8 0
3 years ago
Help me on this pleaseee
lisabon 2012 [21]

Answer:

Bolded below are the answers

Step-by-step explanation:

4568 ÷ 8

(4000 ÷ 8) + (560 ÷ 8) + (8 ÷ 8)

500 + 70 + 1

571

Hope this helps!

8 0
2 years ago
Read 2 more answers
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