Answer:
D. (x – 3)^2 = 17
Step-by-step explanation:
We assume you want an equivalent to x^2 -6x = 8.
We can complete the square by adding the square of half the x-coefficient to both sides of the equation.
x^2 -6x +(-6/2)^2 = 8 +(-6/2)^2
(x -3)^2 = 17 . . . . simplify
Answer:
6. Multiply 2 by 6
7. 71 + 9(8) = 71 + 72 = 143
8. 52 + 3(4) = 12 + 52 = 64
9. 19.25 + 2.75(7) = 19.25 + 19.25 = 38.50
10. 2(0.6)^2 + 4(0.6)(5) = 0.72 + 12 = 12.72
Srry if they r wrong
There are 1000 grams in a kilogram.
9000 grams/1000 grams per kg=9 kg
Finl answer: 9kg
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).
