8/7 as an improper and 1 and 1/7 for proper.
(2y² + 7y + 11) - (8y² - 5y + 7) you can distribute the negative sign to the second expression and then combine the like terms
new expression: (2y² + 7y + 11) (-8y² + 5y - 7)
2y² and -8y² equals -6y²
7y and 5y equals 12y
11 and -7 equals 4
The answer is -6y² + 12y + 4
Answer:
(2 a - 5) (2 a + 5)
Step-by-step explanation:
Factor the following:
4 a^2 - 25
4 a^2 - 25 = (2 a)^2 - 5^2:
(2 a)^2 - 5^2
Factor the difference of two squares. (2 a)^2 - 5^2 = (2 a - 5) (2 a + 5):
Answer: (2 a - 5) (2 a + 5)
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:
1080 m^2 Don't submit m^2 in your answer.
Step-by-step explanation:
Givens
The catch is to find h
To do that, use a^2 + b^2 = c^2
a b and c are in the same 1/2 triangle.
a = 48/2 = 24 m
b = h = ?
c = 51 meters
Solution
a^2 + b^2 = 51^2 Substitute for b^2 = h^2
24^2 + h^2 = 51^2 Expand 24^2 and 51^2
576 + h^2 = 2601 Subtract 576 from both sides
h^2 = 2601 - 576
h^2 = 2025 Take the square root of both sides
h = 45
Area
Area = 1/2 b * h
Area = 1/2 48 * 45
Area = 1080
Remark
Notice that to find h you only use 1/2 of 48 because that is the base of the right triangle.
To find the area, you need to use all of 48 because 48 is the full length of the base.
