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Ksivusya [100]
3 years ago
13

Given this equation, what is the largest possible value for z?

Mathematics
1 answer:
Gre4nikov [31]3 years ago
7 0
Hello,
z=-2x²+15x-3y²+21y-6
=-2(x²-2*15/2*x+(15/2)²) -3(y²-2*7/2y+(7/2)²)-6+2*225/4+3*49/4
=-2(x-15/2)²-3(y-7/2)²+573/4

The max is for x-15/2=0 and y-7/2=0 thus 537/4

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Please help me and explain how you got the answer
PSYCHO15rus [73]

Answer:

∠NQP = 74°

Step-by-step explanation:

NPQ is a triangle.

We know the sum of 3 angles of a triangle is 180 degrees. So we can write:

N + P + Q = 180

2x + 34 + 2x + 2 = 180

Now, we can solve for x:

2x + 34 + 2x + 2 = 180\\4x+36=180\\4x=180-36\\4x=144\\x=\frac{144}{4}\\x=36

The measure of NQP is "2x+2", we plug in x = 36, and find the measure of NQP:

∠NQP = 2(36) + 2 = 74°

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