P = 2(L + W)
P = 400
L = W + 40
400 = 2(W + 40 + W)
400 = 2(2W + 40)
400 = 4W + 80
400 - 80 = 4W
320 = 4W
320/4 = W
80 = W.....so the width (W) = 80 yards
L = W + 40
L = 80 + 40
L = 120...and the length (L) = 120 yards
In around 6.35 years, the population will be 1 million.
<h3> how many years will it take for the population to reach one million?</h3>
The population is modeled by the exponential equation:
Then we just need to solve the equation for t:
Let's solve that:
If we apply the natural logarithm to both sides:
So in around 6.35 years, the population will be 1 million.
If you want to learn more about exponential equations:
brainly.com/question/11832081
#SPJ1
Polynomials are determined to be a trinomial, binomial, etc by the amount of terms they have
For example
7x+3 is a binomial
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).
