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3 years ago

The perimeter of a rectangle is 24 meters. If the length of the rectangle is 8 meters, what is the width (in meters)?

Mathematics

2 answers:

Serhud [2]3 years ago

7 0

Answer:

8 meters

Step-by-step explanation:

8 + 8 + 8 + 8 = 24 which is the perimeter, just trust me

denis-greek [22]3 years ago

6 0

The answer would be 8 meters for sure

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For an investment of $26,245, a quarterly statement reports that the account balance is $26,292. The statement also reports that

Paul [167]

No, it does not seem reasonable. The quarterly statement reports an increase of $47, so it would be a positive rate of return on the investment instead of a negative rate.

6 0

3 years ago

A certain animated movie earned $1.1*10^9 in revenues at the box office. The movie lasts 9.1*10^1 minutes. How much revenue was

Vadim26 [7]

Answer:

$12,087,912.1

Step-by-step explanation:

$1.1*10^9$ is in scientific notation, to get this into normal number, we take the decimal point and move it "9" times to the right [whatever power of 10 we have, we move to right (if positive).] Then we fillup the rest with zeros.

So we have:

$1.1*10^9=1,100,000,000$

This is the revenue.

We do similar process for time:

$9.1*10^1=91$

This is the time , in minutes.

We want revenue PER minute, that means we divide the total revenue by total time, that would be:

$\frac{1,100,000,000}{91}=12,087,912.1$

The revenue earned per minute is $12,087,912.1

6 0

3 years ago

Identify a pattern in each list of numbers. Then use this pattern to find the next number.

Galina-37 [17]

$\begin{array}{cccccccc}\\ \underline{-3}&,&12&,&-33&,&102\\ \cline{1-7} &&(-3)(\underline{-3})+3&&(-3)(12)+3&&(-3)(-33)+3 \end{array} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{ccc} -303&,&912\\ \cline{1-3} (-3)(102)+3&&(-3)(-303)+3 \end{array}$

4 0

2 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$