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Katarina [22]
2 years ago
13

Least value of expression x^2+4y^2+3z^2-4x-20y-12z+45 is.?

Mathematics
1 answer:
Korvikt [17]2 years ago
7 0
Hello,

4 is the least value.

Indeed:
x²-4x+4=(x-2)²
4y²-20y+25=(2y-5)²
3(z²-4z+4)=3(z-2)²

x²+4y²+3z²-4x-20y-12z+45= (x-2)²+(2y-5)²+3(z-2)²+45-4-25-12
=(x-2)²+(2y-5)²+3(z-2)²+4

Minimum is 4 when  x=2,y=5/2 and z=2 for square is always positive.

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If you are using Cosine and need to solve for an unknown hypotenuse, you should...
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<h3 /><h3 /><h3>Trigonometric ratios:</h3>
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Using cosine, the hypotenuse side can be solved as follows:

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