Answer:
=4........,.......,,........,
Answer:
m = 2
Step-by-step explanation:
Slope Formula: 
Simply plug our coordinates into the slope formula and solve for <em>m</em>:
m = (5 - 3)/(4 - 3)
m = 2/1
m = 2
The x-value of the intersection point of the lines is ( d - b)/(a-c)
<h3>Intersection of lines</h3>
Given the equation of a line expressed as;
y = ax +b and;
y = cx + d.
Equating both lines, we will have:
Make x the subject of the formula:
ax - cx = d = b
x (a-c) = d - b
x = ( d - b)/(a-c)
Hence the x-value of the intersection point of the lines is ( d - b)/(a-c)
Learn more on intersecting lines here: brainly.com/question/2065148
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)
Answer:
f(w) = 3w + 2,000,000/w
Step-by-step explanation:
We know that the area of a rectangle is the product of its length and width:
A = LW
Filling in the given values lets us write an expression for the length of the field.
1,000,000 = Lw
L = 1,000,000/w
Since there are 3 fences of length w and two of length L, the total perimeter fence length is the sum ...
f(w) = 3w + 2(1,000,000)/w
Combining the constants, we have a function for the perimeter fence length in terms of the width of the field:
f(w) = 3w +2,000,000/w
