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3 years ago

URGENT! How do you solve a word problem using the direct variation formula?

Mathematics

2 answers:

zhuklara [117]3 years ago

4 0

Divide it
hope this help

Marat540 [252]3 years ago

3 0

Answer:

U DIVIDE IT!!

Step-by-step explanation:

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Which of the expressions are equivalent to the one below? Check all that

VikaD [51]

Answer:

none of them match

Step-by-step explanation:

maybe something is typed wrong. the answer I got for the equation was 108 and none of them match

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2 years ago

Can anyone help this is a hard question.

dolphi86 [110]

The answer is c.......

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2 years ago

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……………………………………………..()1!-!/

alina1380 [7]

Answer:

A. a is where performance is minimal

Step-by-step explanation:

b is representing peak performance.

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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:

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

The sum telescopes so that

$S_n=\dfrac2{14}-\dfrac2{5n+4}$

and as $n\to\infty$ , the second term vanishes and leaves us with

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17$

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3 years ago

“which of the <br> equations below could be the equation of this parabola?”

luda_lava [24]

The answer is C because it is on the x axis and it is positive

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2 years ago

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