Answer: a²+b² = -99/2
Step-by-step explanation:
Since we are given two equations, this equations will be solved simultaneously to get a² and b²
a³ - 3ab² = 47 ... 1
b³ - 3a² b = 52... 2
From 1, a(a² - 3b²) = 47...3
From 2, b(b² - 3a²) = 52... 4
Adding 3 and 4, we have;
a²+b²-3b²-3a² = 99 (note that a and b will no longer be part of the equations as they have been factored out)
a²+b²-(3b²+3a²) = 99
(a²+b²) -3(b²+a²)= 99
Taking the difference we have
- 2(a²+b²) = 99
a²+b² = -99/2
HETY is a parallelogram.
HT and EY are diagonals. We know that diagonals divides the parallelogram into two equal parts.
So ar(HET) = ar(HTY)
And, ar(HEY) = ar(EYT) now, in AHET, diagonal EY bisects the line segment HT and also the AHET,
∴ar(AHOE) = ar(AEOT)
Similarly in AETY
ar(ΔΕΟΤ) = ar(ΔΤΟΥ)
And in AHTY,
ar(ATOY) = ar(AHOY)
That means diagonals in parallelogram divides it into four equal parts.
Hence Proofed.
If R is the midpoint of PS, then PR = RS -- (1)
Also, PR + RS = PS -- (2)
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PR + RS = PS
7x + 23 + 13x - 19 = PS
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Now, PR = PS
7x + 23 = 13x - 19
7x - 13x = -19 - 23
-6x = -42
x = -42/-6
x = 7
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PS = 7x + 23 + 13x - 19
PS = 7(7) + 23 + 13(7) - 19
PS = 49 + 23 + 91 - 19
<u>PS </u><u>=</u><u> </u><u>1</u><u>4</u><u>4</u><u> </u>
Hope it helps!
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