Since we know what he earns hourly, and his sales, we can figure out his full gross pay for the week.
As so:
We can set up an equation like this:
<em>h</em> = hours
<em>s</em> = sales
9.60h + 5%s
Now, we substitute.
9.60(35) + 5%(230)
Next, solve.
336.00 + 11.50
$347.50
$347.50 was Glen's gross pay for the week.
Hope I could help! If my math is wrong, or I did not provide the answer you were looking for, let me know! However, if I did well, please consider marking as <em>Brainliest!
</em>Have a good one!
Answer:
6x^2 - 10x - 24
Step-by-step explanation:
- Do 3x times 2x which is 6x^2
- Do 3x times - 6 which is -18x
- Do 4 times 2x which is 8x
- Do 4 times - 6 which is - 24
- You get 6x^2 - 18x + 8x - 24
- Simplyify to get the answer
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).
Answer:
answer becit uses b
Step-by-step explanation:
but yass queen but annyway
