The slope of AB and the slope of DC based on the information about the coordinate will be -2 and -2 respectively.
<h3>How to illustrate the information?</h3>
From the information given, we are told to use the slope formula along with the coordinates of points A, B, C, and D to find the slope.
The slope of AB based on the information will be:
= (-1 -3)/(-1 --3)
= -4/2
= 2
The slope of DC will be:
= 2-6/(1 --1)
= -4/2
= -2
Therefore, the slope of AB and the slope of DC based on the information about the coordinate will be -2 and -2 respectively.
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Answer:
Step-by-step explanation:
Before we find x, we need to set up this triangle a little more. We need to find the triangle's altitude before we can solve for x. We will use the heartbeat method to find the altitude.
Let altitude = y; solve for y:
Now that we know the altitude, we can use the Pythagorean Theorem to find the hypotenuse (x).
Since c and x are the same; c is just the hypotenuse in the Pythagorean Theorem.
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).
y = 2 + x^6
y - 2 = x^6
x = (y - 2)^1/6
Well, solving this integral
= 2,69279
