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2 years ago

Round 0.01123 to the nearest hundredth?

Mathematics

2 answers:

Rasek [7]2 years ago

7 0

The answer to your question would be 0.01

deff fn [24]2 years ago

5 0

The answer is 0.01 going by decimal places. Decimal places always beging with tenths then proceed to hundredths, thousandths and so forth

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OMG PLEASE Help me !!!! ILL GIVE YOU BRAINLIEST !!! Explain in order to get it

Doss [256]

Answer:

I'm not sure but I believe the correct answer is line L2.

Step-by-step explanation:

It seems to be closer to more points than the other lines.

If it is not L2, it must be L3.

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2 years ago

ZX and zy or supplementary angles. zy measures 49 degrees what is the measure of ZX

Svetach [21]

Answer:

131˙

Step-by-step explanation:

Supplementary angles are angles that add up to 180˙. If you subtract 49 from 180, you get 131.

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2 years ago

Which property or properties of arithmetic do you use when you calculate 3×70 by first calculating 3×7 = 21 and then putting a z

ANEK [815]

The property used is the Associative property

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The formula is

a × bc = a (bc) = a (b×c) = (a×b) ×c

3 0

3 years ago

Find the slope of the line: (4, 4), (5, 3)

fredd [130]

Answer:

-1

Step-by-step explanation:

we would use the slope formula which is y2-y2/x2-x1

so 3-4/5-4= -1

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2 years ago

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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