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3 years ago

14

Consider the following 8 numbers, where one labelled X is unknown.

1 answer:

Effectus [21]3 years ago

7 0

Answer:

$55,-3$

Step-by-step explanation:

$We\ are\ given\ that,\\There\ are\ eight\ numbers\ in\ the\ data :\\17,18,24,x,44,48,42,4\\Where\ the\ value\ of\ x\ is\ what\ we\ need\ to\ find.\\Range\ of\ the\ data=51\\Hence,\\We\ know\ that\ the\ range\ is\ the\ difference\\ between\ the\ highest\ observation\ and\ the\ lowest\ observation\ in\ the\ data.\\We\ may\ know\ see\ two\ cases\ of\ evaluating\ x\ when:\\1. x\ is\ the\ highest\ observation\ in\ the\ data\ or\ x>48.\\Highest\ observation\ =x\\Lowest\ observation=4\\Hence,\\$

$As\ range=51,\\x-4=51\\x=51+4\ [Adding\ 4\ on\ both\ the\ sides]\\x=55 \\2. x\ is\ the\ lowest\ observation\ or\ x$

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Answer:

The factors of $2(x+y)^2-9(x+y)-5$ is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=> $2(x+y)^2-9(x+y)-5$

To Find:

The factors of the polynomial =?

Solution:

Lets assume k = (x+y)

Then $2(x+y)^2-9(x+y)-5$ can be written as $2k^2-9k-5$

Now by using quadratic formula

k = $\frac{-b\pm\sqrt{(b^2-4ac}}{2a}$

where

a= 2

b= -9

c= -5

Substituting the values, we get

k = $\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}$

k = $\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}$

k = $\frac{-(-9) \pm \sqrt{(81+40)}}{4}$

k = $\frac{-(-9) \pm \sqrt{(121)}}{4}$

k = $\frac{-(-9) \pm 11}}{4}$

k= $\frac{ 9 \pm 11}{4}$

k = $\frac{20}{4}$ k = $\frac{-2}{4}$

$k_1 =5$ $k_2 = -\frac{1}{2}$

$2k^2-9k-5= 2(k-5)(k+\frac{1}{2})$

Solving the RHS we get

$\frac{2}{2}(k-5)(2k+1)$

$(k-5)(2k+1)$

Substituting k = x+y

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Answer:

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Step-by-step explanation:

Check answer:

14+12= 26

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