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Fudgin [204]
2 years ago
6

Solve the system below, choose the x and y answers from the list below.

Mathematics
1 answer:
Bumek [7]2 years ago
5 0

Answer:

y=2, x=-1

Step-by-step explanation:

We can add the two equations together to eliminate x:

6x+5y+(-6x)+7y=4+20

12y=24

y=2

We plug y into the first equation and get

6x+5(2)=4

6x+10=4

6x=-6

x=-1

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Sonbull [250]
The numerator and denominator can be divided by 4 to get 12/25.

48 / 4 = 12
100 / 4 = 25
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3 years ago
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Jlenok [28]
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3 . Find the dimensions of the box that requires the least amount of cardboard. Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From these, we obtain x 2y = 8 = xy2 . This forces x = y = 2, which forces z = 1. Calculating second derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus, moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
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3 years ago
Maria mows 2/3 of a lawn in 1/5 hour. How Many similar-sized lawns can she mow in 1 hour?
sp2606 [1]

Answer:

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2/3 of a lawn in 1/5 hour

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