To answer this, let's first describe the two areas and obtain the pertinent dimensions from them.
The area of the square hole is 5 cm^2. Since A = s^2, where s is the length of a side of the square, s = +√5 in this situation. +√5 is approx. 2.24 cm.
The area of the round peg is 5 cm^2 also, but the area is calculated using a different formula: A = πr^2, where r is the radius of the circle. Solving for r^2, we get:
r^2 = A/π. Here, r^2 = (5 cm^2)/π = 5π, so that:
r = +√(5π). This is approx. 3.96 cm, and so the diameter is twice that, or 7.93 cm.
So there's plenty of room for the round peg to enter the square hole, but not the other way around!
Let w= number of weeks
400-40w<= 160
btw <= means less than or equal too
Step-by-step explanation:
F=ma
m=f/a
m=200÷5
m=40
bearing in mind that standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient
![\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-7=1[x-(-1)]\implies y-7=x+1 \\\\\\ y=x+8\implies \boxed{-x+y=8}\implies \stackrel{\textit{standard form}}{x-y=-8}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-7%3D1%5Bx-%28-1%29%5D%5Cimplies%20y-7%3Dx%2B1%20%5C%5C%5C%5C%5C%5C%20y%3Dx%2B8%5Cimplies%20%5Cboxed%7B-x%2By%3D8%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bstandard%20form%7D%7D%7Bx-y%3D-8%7D)
just to point something out, is none of the options, however -x + y = 8, is one, though improper.
