Which identity is the result of using the Pythagorean theorem to show that a triangle with side lengths

Mathematics

1 answer:

disa [49]3 years ago

4 0

C I think but I’m not sure

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PPLEASE HELP!!!!! WHAT IS THE INITIAL VALUE MEAN FOR THIS GRAPH/FUNCTION

Rudiy27

Amount saved:50I60I70l80l AND ON......
......weeks......:0..I1...l2..l3..l ..........AND ON
amount saved =x
weeks=y
after every y+10=x

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3 years ago

O) x² – 18x - 19 + 20y - y2,

pshichka [43]

Answer:

19x-19+23y

Step-by-step explanation:

add the like terms

7 0

3 years ago

Which could be the length of Side KL?

Kay [80]

I don’t know the answer to that one sorry

6 0

2 years ago

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Find the sum. 2/5 + 1/2

dybincka [34]

2/5 + 1/2

4+5 / 10 • look for the LCD of 5 and 2
which is 10 then 5 goes 2
times into 10 the multiply by 2
= 4. Then 2 goes 5 times into
10 then multiply by 1 = 5.

Therefore the sum of 2/5 + 1/2 =9/10

5 0

2 years ago

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The savings account offering which of these APRs and compounding periods offers the best APY?

zzz [600]

$\bf \qquad \qquad \textit{Annual Yield Formula}
\\\\
~~~~~~~~~~~~\textit{4.0784\% compounded monthly}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0784\%\to \frac{4.0784}{100}\to &0.040784\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{monthly, thus twelve}
\end{array}\to &12
\end{cases}
\\\\\\
\left(1+\frac{0.040784}{12}\right)^{12}-1\\\\
-------------------------------\\\\
~~~~~~~~~~~~\textit{4.0798\% compounded semiannually}\\\\
$

$\bf ~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0798\%\to \frac{4.0798}{100}\to &0.040798\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2
\end{cases}
\\\\\\
\left(1+\frac{0.040798}{2}\right)^{2}-1\\\\
-------------------------------\\\\
$

$\bf ~~~~~~~~~~~~\textit{4.0730\% compounded daily}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0730\%\to \frac{4.0730}{100}\to &0.040730\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{daily, thus 365}
\end{array}\to &365
\end{cases}
\\\\\\
\left(1+\frac{0.040730}{365}\right)^{365}-1$

7 0

2 years ago

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