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2 years ago

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I am a 3-digit number. If rounded off to the nearest hundreds. I become 500. If rounded off to the nearest tens, I become 540. I

f the sum of my digit 13, what number am I?

1 answer:

dimulka [17.4K]2 years ago

6 0

Answer:

535 is the cirrext answer

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A person invests $4000 at 2% interest compounded annually for 4 years and then invests the balance (the $4000 plus the interest

faltersainse [42]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
A=4000\left(1+\frac{0.02}{1}\right)^{1\cdot 4}\implies A=4000(1.02)^4\implies A\approx 4329.73$

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396$

add both amounts, and that's her investment for the 11 years.

7 0

2 years ago

HELP ME PLEASE I AM BEGGING

Inessa05 [86]

Answer:

H

Step-by-step explanation:

well it would be a proportion. so 5 cans per 3 days gives you 5/3. then for the next proportion. it would also be cans over days. so c / 7

5/3 = c/7

answer is H

8 0

2 years ago

Someone please HELP!?

jeka94

A negative exponent just means that the base is on the wrong side of the fraction line, so you need to flip the base to the other side.

For example

$x^{-2}=\frac{1}{x^2}$

or

$( \frac{2}{5} )^{-4}=( \frac{5}{2} )^4$

***************************************
$(8r^{-5})^{-3}$

$(8* \frac{1}{r^5} )^{-3} \\\\( \frac{8}{r^5} )^{-3}
 \\ \\ ( \frac{r^5}{8} )^3
\\\\ \mathrm{Apply\:exponent\:rule}:\quad *\left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} \ \ \ \ \ \ \ \ \ \ \ \ *(a^b)^c=a^{bc}
 \\\\( \frac{r^{15}}{8^3} )
\\\\( \frac{r^{15}}{512} )$

The answer is "D"

7 0

2 years ago

What is the simplified form of this expression?<br><br> (-3x2 + 2x − 4) + (4x2 + 5x + 9)

olga2289 [7]

Answer:

<em><u>I</u></em><em><u> </u></em><em><u>hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>helps</u></em><em><u> </u></em><em><u>you</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

3 0

2 years ago

Read 2 more answers

The shortest side of a fight triangle measures20 m. The lengths of the other two sides are censecutive odd integers. Find the le

slamgirl [31]

A² + b² = c²
20² + n² = (n+2)²
400+ n² = (n+2)(n+2)
400 + n²=n² + 2n + 2n +4
400 + n² = n² +4n +4
400 + n²-n² = n²-n² + 4n+4
400 = 4n + 4
400-4 = 4n +4-4
396 = 4n
396/4 = 4n/4
n= 99

20²+ 99² = (99+2)²
400 + 9801 =10201
10201=10201

6 0

2 years ago