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ratelena [41]
3 years ago
6

Write an equation to describe this situation then solve it. Eight less than three times a number equals twenty-two

Mathematics
1 answer:
EastWind [94]3 years ago
6 0

Answer:

The number is 10

Step-by-step explanation:

3x-8=22

3x=30

x=10

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The perimeter of the triangle is 13 inches. What is the length of the shortest side?
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Answer:

3 in

Step-by-step explanation:

Sum of all sides of ∆ = perimeter

Given that it has a perimeter of 13 in, and the following sides: (x - 5) in, x/2 in, and 6 in, therefore:

x - 5 + \frac{x}{2} + 6 = 13

Solve to find the value of x

\frac{2x - 10 + x + 12}{2} = 13

\frac{3x + 2}{2} = 13

Multiply both sides by 2

\frac{3x + 2}{2}*2 = 13*2

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Divide both sides by 3

\frac{3x}{3} = \frac{24}{3}

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3 years ago
Consider 8x2 - 48x = -104.
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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
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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
4 years ago
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