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3 years ago

Find the product of the following

Mathematics

1 answer:

wlad13 [49]3 years ago

3 0

Here are your in-depth answers!

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If the dimensions of the following cylinder are doubled, what factor will the surface area change by?

Vedmedyk [2.9K]

Answer:

probably 2 or 4

Step-by-step explanation:

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3 years ago

If master chief has three and a half cookies, and the arbiter has five and three quarters of a cookie, how many cookies do they

nikklg [1K]

Answer:

$9\dfrac{1}{4}$

$4\dfrac{5}{8}$

Step-by-step explanation:

The master chief has three and a half cookies i.e. $3\dfrac{1}{2} = \dfrac{7}{2}$ numbers of cookies.

Again the arbiter has five and three quarters of a cookie i.e. $5\dfrac{3}{4} = \dfrac{23}{4}$ number of cookies.

Therefore, in total there are $(\dfrac{7}{2} +\dfrac{23}{4} ) = \dfrac{37}{4} = 9\dfrac{1}{4}$ numbers of cookies.

If we divide the total number of cookies in to equal parts and distribute to then then each of them will get $\dfrac{37}{4 \times 2} = \dfrac{37}{8} = 4\dfrac{5}{8}$ numbers of cookies. (Answer)

3 0

3 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:

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