3 years ago

20 points: please help me

Mathematics

1 answer:

ki77a [65]3 years ago

3 0

Answer:

0...because before it doesn't have an x2 term....

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NEED HELP QUICK!! :)

Vlada [557]

Let x and y be your two consecutive whole numbers

x < sqrt(142) < y
x^2 < 142 < y

So, we are looking for x and y such that x^2 < 142 and y^2 > 142.

The closest squared number to 142 is 144 = 12^2.
Next is 11^2 = 121.

11 and 12 are consecutive.

11^2 = 121 < 142 < 144 = 12^2

Thus, 11 and 12 are your numbers

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2 years ago

Which one is it please help meee

iragen [17]

Answer:

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P/2 = 3/4 + p/3 solve for p<br> Show your work<br> It is URGENT

il63 [147K]

Step-by-step explanation:

p/2 = 3/4 + p/3

p/2 - p/3 = 3/4

(3p - 2p )/6 = 3/4

p/6 = 3/4

multiply both sides by 6

p = 3×6/4

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2 years ago

3x+3y+2(2x-y) equivalent fractions to this problem

Anastaziya [24]

Distribute the 2 into the 2x and -y

Then add “like terms” which are 3x and 4x... 3y and -2y

Which then becomes 7x+y

There are no more “like terms” so that is your final answer

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2 years ago

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NikAS [45]

I think any of these would work

10;18

15;28

20;35

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2 years ago

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