Answer:
c) 5√2 cm
Step-by-step explanation:
A square with side length l has a perimeter given by the following equation:
P = 4l.
In this question:
P = 20
So the side length is:
4l = 20
l = 20/4
l = 5
Diagonal
The diagonal forms a right triangle with two sides, in which the diagonal is the hypothenuse. Applying the pytagoras theorem.
Lenght is a positive meausre, so
So the correct answer is:
c) 5√2 cm
Slope = Change in Y / change in X
Slope = (8 - 2) / (9-6)
Slope = 6 / 3
Slope = 2
Answer:
c)A cylinder with a volume of π(w³ +2w² )in³
where w represents the width of the cylinder.
Step-by-step explanation:
A cylinder with a volume of π(w³ +2w² )in³
where w represents the width of the cylinder.
The volume of a cylinder is given by V=Bh or V=πr²h
where B is the area of the base and h is the height of the cylinder.
V=πr²h-------equation1
V= π(w³ +2w² )--------- equation2
Comparing equation 1 and equation 2
πr²h=π(w³ +2w² )
r²h=(w³ +2w² )
r²h= w²(w +2 )
Taking w² common leaves w+2 which gives the height or length of the cylinder.
Therefore radius = width and height = length which gives the volume of the cylinder.
first off, let's split the triplet into two equations, then from there on we'll do substitution.
![\cfrac{y}{x-z}=\cfrac{x}{y}=\cfrac{x+y}{z}\implies \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=\cfrac{x+y}{z} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{\cfrac{y}{x-z}=\cfrac{x}{y}\implies }y^2=\underline{x^2-xz} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 2nd equation}}{\cfrac{x}{y}=\cfrac{x+y}{z}\implies }xz=xy+y^2\implies \stackrel{\textit{substituting for }y^2}{xz=xy+(\underline{x^2-xz})}](https://tex.z-dn.net/?f=%5Ccfrac%7By%7D%7Bx-z%7D%3D%5Ccfrac%7Bx%7D%7By%7D%3D%5Ccfrac%7Bx%2By%7D%7Bz%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Ccfrac%7By%7D%7Bx-z%7D%3D%5Ccfrac%7Bx%7D%7By%7D%5C%5C%5B2em%5D%20%5Ccfrac%7Bx%7D%7By%7D%3D%5Ccfrac%7Bx%2By%7D%7Bz%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20the%201st%20equation%7D%7D%7B%5Ccfrac%7By%7D%7Bx-z%7D%3D%5Ccfrac%7Bx%7D%7By%7D%5Cimplies%20%7Dy%5E2%3D%5Cunderline%7Bx%5E2-xz%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20the%202nd%20equation%7D%7D%7B%5Ccfrac%7Bx%7D%7By%7D%3D%5Ccfrac%7Bx%2By%7D%7Bz%7D%5Cimplies%20%7Dxz%3Dxy%2By%5E2%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20for%20%7Dy%5E2%7D%7Bxz%3Dxy%2B%28%5Cunderline%7Bx%5E2-xz%7D%29%7D)
![2xz=xy+x^2\implies 2xz=x(y+x)\implies \cfrac{2xz}{x}=y+x \\\\\\ 2z=y+x\implies 2=\cfrac{y+x}{z}\implies 2=\cfrac{x+y}{z} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{}{ \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=\cfrac{x+y}{z} \end{cases}}\implies \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=2 \end{cases}\implies \begin{cases} \cfrac{y}{x-z}=2\\[2em] \cfrac{x}{y}=2 \end{cases}](https://tex.z-dn.net/?f=2xz%3Dxy%2Bx%5E2%5Cimplies%202xz%3Dx%28y%2Bx%29%5Cimplies%20%5Ccfrac%7B2xz%7D%7Bx%7D%3Dy%2Bx%20%5C%5C%5C%5C%5C%5C%202z%3Dy%2Bx%5Cimplies%202%3D%5Ccfrac%7By%2Bx%7D%7Bz%7D%5Cimplies%202%3D%5Ccfrac%7Bx%2By%7D%7Bz%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%7D%7B%20%5Cbegin%7Bcases%7D%20%5Ccfrac%7By%7D%7Bx-z%7D%3D%5Ccfrac%7Bx%7D%7By%7D%5C%5C%5B2em%5D%20%5Ccfrac%7Bx%7D%7By%7D%3D%5Ccfrac%7Bx%2By%7D%7Bz%7D%20%5Cend%7Bcases%7D%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Ccfrac%7By%7D%7Bx-z%7D%3D%5Ccfrac%7Bx%7D%7By%7D%5C%5C%5B2em%5D%20%5Ccfrac%7Bx%7D%7By%7D%3D2%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Ccfrac%7By%7D%7Bx-z%7D%3D2%5C%5C%5B2em%5D%20%5Ccfrac%7Bx%7D%7By%7D%3D2%20%5Cend%7Bcases%7D)
that of course, is only true if x + y, or our numerator doesn't turn into 0, if it does then our fraction becomes 0 and our equation goes south. Keeping in mind that x,y and z are numeric values that correlate like so.
