Answer:
c
Step-by-step explanation:
A perfect square trinomial is always in the form
(a+b)(a+b)= 
ie. (x+4)(x+4)= 
This would be True. For example. If your number is 24, and 'n' is 12 ( a factor of 24 )
your factors of 12 (12, 6, 4, 3, 2, 1) are also included in the factors of 24 (which are 24, 12, 6, 4, 3, 2, 1)
Answer:
30 I am pretty sure
Step-by-step explanation:
First, to make things easier, let's turn 5/4 into a decimal given a second is equal to 1000 milliseconds.
5/4 = 1,25
1,25 * 1000 = 1250 milliseconds = 1.25 seconds
So, we've found that 45 yards can be covered in 1.25 seconds. Now we'll use the rule of 3:
45 - 1.25
80 - x
Now we solve:
(80 * 1.25) / 45 = x
100 / 45 = x
2,22... = x
So, according to our answer, 80 yards can be covered in 2,22 seconds, which in fractions would be:
Hope it helped,
BioTeacher101
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).
