Answer:
j / 9 = 5
j = 45
Step-by-step explanation:
j / 9 = 5
Multiply each side by 9
j/9 * 9 = 5*9
j = 45
The answer is 1.02 according to calculations
Answer:
y = 55x + 15
Step-by-step explanation:
i think this is it, but it might not be right
<span>Length = 1200, width = 600
First, let's create an equation for the area based upon the length. Since we have a total of 2400 feet of fence and only need to fence three sides of the region, we can define the width based upon the length as:
W = (2400 - L)/2
And area is:
A = LW
Substitute the equation for width, giving:
A = LW
A = L(2400 - L)/2
And expand:
A = (2400L - L^2)/2
A = 1200L - (1/2)L^2
Now the easiest way of solving for the maximum area is to calculate the first derivative of the expression above, and solve for where it's value is 0. But since this is supposedly a high school problem, and the expression we have is a simple quadratic equation, we can solve it without using any calculus. Let's first use the quadratic formula with A=-1/2, B=1200, and C=0 and get the 2 roots which are 0 and 2400. Then we'll pick a point midway between those two which is (0 + 2400)/2 = 1200. And that should be your answer. But let's verify that by using the value (1200+e) and expand the equation to see what happens:
A = 1200L - (1/2)L^2
A = 1200(1200+e) - (1/2)(1200+e)^2
A = 1440000+1200e - (1/2)(1440000 + 2400e + e^2)
A = 1440000+1200e - (720000 + 1200e + (1/2)e^2)
A = 1440000+1200e - 720000 - 1200e - (1/2)e^2
A = 720000 - (1/2)e^2
And notice that the only e terms is -(1/2)e^2. ANY non-zero value of e will cause this term to be non-zero and negative meaning that the total area will be reduced. Therefore the value of 1200 for the length is the best possible length that will get the maximum possible area.</span>
well, if the diameter is 5, thus its radius must be half that, or 2.5, and therefore, the radius of the one four times as much will be (4)(2.5).
Let's simply get their difference, since that'd be how much more is needed from the smaller to larger sphere.
![~\hfill \stackrel{\textit{surface area of a sphere}}{SA=4\pi r^2}\qquad \qquad r=radius~\hfill \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large difference of their areas}}{\stackrel{\textit{radius of (4)(2.5)}}{4\pi (4)(2.5)^2}~~ - ~~\stackrel{\textit{radius of 2.5}}{4\pi (2.5)^2}}\implies 100\pi -25\pi \implies 75\pi ~~ \approx ~~235.62~ft^2](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsurface%20area%20of%20a%20sphere%7D%7D%7BSA%3D4%5Cpi%20r%5E2%7D%5Cqquad%20%5Cqquad%20r%3Dradius~%5Chfill%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20difference%20of%20their%20areas%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Bradius%20of%20%284%29%282.5%29%7D%7D%7B4%5Cpi%20%284%29%282.5%29%5E2%7D~~%20-%20~~%5Cstackrel%7B%5Ctextit%7Bradius%20of%202.5%7D%7D%7B4%5Cpi%20%282.5%29%5E2%7D%7D%5Cimplies%20100%5Cpi%20-25%5Cpi%20%5Cimplies%2075%5Cpi%20~~%20%5Capprox%20~~235.62~ft%5E2)
