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Semmy [17]
3 years ago
15

If (x-y)^2=100 and xy=20. what's the value of x^2+y^2?

Mathematics
2 answers:
ahrayia [7]3 years ago
8 0

(x-y)^2=x^2-2xy+y^2=(x^2+y^2)-2(xy)\\\\\text{We have}\ (x-y)^2=100\ \text{and}\ xy=20.\\\\\text{Substitute}\\\\100=(x^2+y^2)-2(20)\\100=(x^2+y^2)-40\qquad\text{add 40 to both sides}\\\\140=(x^2+y^2)\\\\Answer:\ \boxed{x^2+y^2=140}

Flauer [41]3 years ago
7 0

<u>Answer:</u> The value of x^2+y^2 is 140

<u>Step-by-step explanation:</u>

We are given an expression:

(x-y)^2=100

And,

xy = 20

Using the identity:

(a-b)^2=a^2+b^2-2ab

Solving the expression:

x^2+y^2-2xy=100

Putting the value of 'xy' in above equation, we get:

x^2+y^2-2(20)=100\\\\x^2+y^2=100+40\\\\x^2+y^2=140

Hence, the value of x^2+y^2 is 140

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8 0
3 years ago
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Find the sum of the first one hundred positive integers. see fig 6.26
nignag [31]
Here's a pattern to consider:
1+100=101
2+99=101
3+98=101
4+97=101
5+96=101
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This question relates to the discovery of Gauss, a mathematician. He found out that if you split 100 from 1-50 and 51-100, you could add them from each end to get a sum of 101. As there are 50 sets of addition, then the total is 50×101=5050
So, the sum of the first 100 positive integers is 5050.

Quick note
We can use a formula to find out the sum of an arithmetic series:
s =  \frac{n(n + 1)}{2}
Where s is the sum of the series and n is the number of terms in the series. It works for the above problem.
8 0
3 years ago
Seth has one less than twice the number of compact discs (CDs) that Jason has. Raoul has 53 more CDs than Jason. If Seth gives J
ELEN [110]
S = 2J - 1
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S - 25 = J + 25...S = J + 50

2J - 1 = J + 50
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J = 51 <===Jason has 51 CD's

R = J + 53
R = 51 + 53
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S = 2J - 51
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4 0
4 years ago
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kenny6666 [7]

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If this is confusing, I can explain further. :D

5 0
3 years ago
Read 2 more answers
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