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3 years ago

Find the number that makes the ratio equivalent to 5:4. :48

Mathematics

1 answer:

Orlov [11]3 years ago

5 0

Answer:

Step-by-step explanation:

5:4 = 60:48

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Which expression is equivalent to the following complex fraction? 1-1/x/2

pishuonlain [190]

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3 years ago

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Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the

andrew-mc [135]

Answer:

Wheres the picture?

Step-by-step explanation:

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3 years ago

Suppose that you have $6000 to invest. Which investment yields the greater return over four years: 8.25% compounded quarterly or

WARRIOR [948]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$6000\\
r=rate\to 8.25\%\to \frac{8.25}{100}\to &0.0825\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &4
\end{cases}
\\\\\\
A=6000\left(1+\frac{0.0825}{4}\right)^{4\cdot 4}\implies A=6000(1.020625)^{16}\\\\
-------------------------------\\\\$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$6000\\
r=rate\to 8.3\%\to \frac{8.3}{100}\to &0.083\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semiannually, thus two}
\end{array}\to &2\\
t=years\to &4
\end{cases}
\\\\\\
A=6000\left(1+\frac{0.083}{2}\right)^{2\cdot 4}\implies A=6000(1.0415)^8$

compare them away.

4 0

3 years ago

The sector of a circle with a 12-inch radius has a central angle measure of 60°. What is the exact area of the sector in terms o

goldfiish [28.3K]

75.3982237 inches squared is the answer, round it to whatever decimal place your asked to.

3 0

3 years ago

What is a rule for

USPshnik [31]

Answer:

Step-by-step explanation:f(h(x))= 2x -21

f(x)= x^3 - 6

h(x)=\sqrt[3]{2x-15}

WE need to find f(h(x)), use composition of functions

Plug in h(x)

f(h(x))=f(\sqrt[3]{2x-15})

Now we plug in f(x) in f(x)

f(h(x))=f(\sqrt[3]{2x-15})=(\sqrt[3]{2x-15})^3 - 6

cube and cube root will get cancelled

f(h(x))= 2x-15 -6= 2 x-21

7 0

2 years ago