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creativ13 [48]
3 years ago
14

McKenzie and Cara both tried to find the missing side of the right triangle. Explain your reasoning. Please explain why one of t

hem is correct

Mathematics
2 answers:
Paladinen [302]3 years ago
7 0
By the Pythagorean Theorem:

h^2=x^2+y^2

Which just means that the hypotenuse (longest side) squared is equal to the sum of the side lengths squared...in this case:

h^2=5^2+13^2

h^2=25+169

h^2=194

h=√194

h≈13.93 in (to nearest hundredth of an inch)

Mckensey is correct as shown above.

Cara's mistake was in letting c be a side length instead of the length of the hypotenuse...
Andreas93 [3]3 years ago
3 0

Answer:

Cara's mistake was in letting c be a leg length instead of the length of the hypotenuse.

Step-by-step explanation:

chart shows documented error.

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<span>The probability that a house in an urban area will develop a leak is 55​%. if 20 houses are randomly​ selected, what is the probability that none of the houses will develop a​ leak? round to the nearest thousandth.

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3 years ago
Suzanne lays the books end to end and finds that the board is the same length as 21 bucks. How many centimeters long is the boar
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3 years ago
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vredina [299]

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Step-by-step explanation:

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A lidless box is to be made using 2m^2 of cardboard find the dimensions of the box that requires the least amount of cardboard
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).
5 0
3 years ago
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