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3 years ago

Is 5622 divisible by 4

Mathematics

2 answers:

Pani-rosa [81]3 years ago

7 0

Well it's 1,405.5 so no

frez [133]3 years ago

3 0

Actually no because when we divide 5622/4 = 2811/2. So it is not divisible.

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A soccer ball was kicked in the air and follows the path h(x)=−2x2+1x+6, where x is the time in seconds and h is the height of t

madam [21]

The ball will hit the ground at 2 seconds.

Step-by-step explanation:

Given that,

The path of the ball = h(x)=−2x2+1x+6

Here,

x is the time while h is the height of ball.

When the ball will hit the ground, the height will become zero. Therefore,

h(x)=−2x2+1x+6

0 =−2x2+1x+6

2x2 -1x - 6 = 0

This is a quadratic equation, hence by applying quadratic equation formula:

$x = \frac{-b +- \sqrt{b^{2} - 4ac } }{2a}$

here,

a = 2

b = -1

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Putting these values in formula, we get

$x = \frac{-(-1) +- \sqrt{-1^{2} - 4.2.-6 } }{2.2}$

$x = \frac{1 +- \sqrt{1 + 48 } }{4}$

$x = \frac{1 +- \sqrt{49 } }{4}$

$x = \frac{1 +- 7 }{4}$

x = 2, -3/2

As the time cannot be negative. Therefore, the ball will hit the ground at 2 seconds.

6 0

3 years ago

Read 2 more answers

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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Ainat [17]

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uysha [10]

Answer:

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Step-by-step explanation:

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