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3 years ago

Please answer correctly !!!!!!!! Will mark brianliest !!!!!!!!!!!

Mathematics

1 answer:

nexus9112 [7]3 years ago

4 0

Answer:

x=6

Step-by-step explanation:

h(x) = -( x-2)^2 +16

We want when h(x) = 0

0 = -( x-2)^2 +16

Subtract 16 from each side

-16 = -( x-2)^2 +16-16

-16 = -( x-2)^2

Divide by -1

16= ( x-2)^2

Take the square root of each side

±sqrt(16) = sqrt(( x-2)^2 )

±4 = x-2

Add 2 to each sdie

2 ±4 = x-2+2

2+4 = x 2-4 =x

6 =x -2 =x

since time cannot be negative

x=6

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3 years ago

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4 years ago

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3 years ago

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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:

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

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If $k=\frac15$ , then

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This means

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

The sum telescopes so that

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3 years ago

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3 years ago