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erastova [34]
3 years ago
5

Help me plspslsplpsl 42x^2+34-8-12

Mathematics
1 answer:
OverLord2011 [107]3 years ago
8 0

\bf Step-by-step~explanation:

\bf Step~1:

To solve this, we have to simplify these numbers by adding like terms.

42x^2+34-8-12\\\\(34-8=26, 26-12=14)\\\\42x^2+14\\

\bf Step~2:

Next, we factor our expression. To do so, we look for the greatest common factor of 42 and 14, which is 14. We know that because 42 divided by 14 is 3.

\frac{42}{14} =3

\boxed{ \bf Our~final~answer: 14(3x^2 + 1)}

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x > 2 ---- Graphed with an open circle, everything greater than but not including 2

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3 years ago
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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).
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3 years ago
Find the x intercepts of the parabola with vertex (-1,-16) and y intercept (0,-15)
Arlecino [84]
<span> first, write the equation of the parabola in the required form: </span>
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<span>Here, (h, k) is given as (-1, -16). </span>
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<span>Or, if you prefer, (3, 0), (-5, 0). </span>
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