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maria [59]
3 years ago
10

If 2a+3b=12 and ab=6 find the value of 8a^3+27b^3

Mathematics
2 answers:
lisov135 [29]3 years ago
8 0
A b = 6
a = 6/b
2 *6/b + 3b = 12
12/b + 3b = 12
12 + 3b^2 = 12b
3b^2 - 12b + 12 = 0
b^2 - 4b +4 = 0
(b -  2)^2 = 0
b = 2
a will = 3
8*3^3 + 27*2^3
8*27 + 27*8
I'm getting 432

I thought perhaps I could do this a shorter way. Turns out I can't. If you give out a Brainly Please give it to the other responder.
Alex17521 [72]3 years ago
6 0
<span>A) 2a + 3b = 12
B) ab = 6 solving for a
B) a = 6 / b then we substitute this into equation A)
</span><span>A) 12 / b + 3b = 12  </span><span>multiplying this by "b"
A) 12 + 3b^2 = 12b
A) 3b^2 -12b +12= 0 dividing by "3"
A) b^2 -4b + 4 = 0
Factoring
(b-2) * (b-2) = 0
b = 2
Since b = 2 then a = 3

</span>NOW, we put these numbers into:
<span>8a^3 +27b^3 </span>
8*3*3*3 + 27*2*2*2
216 + 216
The answer is 512


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cricket20 [7]

Answer:

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Obviously, all multiples of 3 and 7 can be paid.

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3 years ago
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docker41 [41]
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Given:

The given sets are:

Set a : 200, 104, 100, 160.

Set b: 270, 400, 483, 300, x.

Mean of set a: mean of set b= 3:8

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Mean=\dfrac{270+400+483+300+x}{5}

Mean=\dfrac{1453+x}{5}

Mean=\dfrac{1453+x}{5}

It is given that,

Mean of set a: mean of set b= 3:8

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