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wel
3 years ago
8

If a varies inversely as the cube root of b and a=1 when b=64, find b​

Mathematics
1 answer:
Mrrafil [7]3 years ago
7 0

Answer:

  b = 64/a³

Step-by-step explanation:

Using the given information, we can only find a relation between a and b. We cannot find any specific value for b.

Since a varies inversely as the cube root of b, we have ...

  a = k/∛b

Multiplying by ∛b lets us find the value of k:

  k = a·∛b = 1·∛64 = 4

Taking the cube of this equation gives ...

  64 = a³b

  b = 64/a³ . . . . . divide by a³

The value of b is ...

  b = 64/a³

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I'm guessing the sum is supposed to be

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Split the summand into partial fractions:

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

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If k=-\frac45, then

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If k=\frac15, then

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This means

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

Consider the nth 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}

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

7 0
3 years ago
PLEASE HELP!!!!!!
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B

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To multiply fractions use the method
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6 0
3 years ago
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