Lateral Area of a cylinder is ; 2πR.H
R=12 mm and H= 15 mm
Then LATERAL area = 2π(12)(15) =48π ≈ 150 mm²
31 quarts, because there are 4 quarts in a gallon
we'll start off by grouping some

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?
well, let's recall that a perfect square trinomial is

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²
![\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29%3D6%5Cimplies%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D6%2B%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D%5Ccfrac%7B25%7D%7B4%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Csqrt%7B%5Ccfrac%7B25%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B%5Csqrt%7B25%7D%7D%7B%5Csqrt%7B4%7D%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B5%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B6%7D%7B2%7D%5Cimplies%20%5Cboxed%7Bx%3D3%7D)
Answer:
triangle AEH, triangle CFG, triangle BFG, and triangle DEG.
so b,c,e,f
this is right for e2020.
So we are given two points that the line crosses, the origin and (9, -3), we can calculate the slope m of the line with these data, dividing the y segment by the x segment:
m = (-3 - 0)/(9 - 0) = -3/9
m = -1/3
then we can use the point slope line equation to find the line equation, lets use the point origin (0,0) to do so:
y - y1 = m(x - x1), where x1, y1 are the coordinates of a point that the line crosses:
y - 0 = (-1/3)(x - 0)
y = <span>(-1/3)x
so this is the equation of the line, slope -1/3 and y intercept 0</span>