First find the tangent line
dy/dx=2x
at x=6, the slope is 2(6)=12
so
use point slope form
y-y1=m(x-x1)
point is (6,36)
so
y-36=12(x-6)
y-36=12x-72
y=12x-36
alright, so we know they intersect at x=6
and y=12x-36 is below y=x^2
so we do
![[\frac{x^3}{3}-6x^2+36x]\limits^6_0=(\frac{6^3}{3}-6(6)^2+36(6))-(0)=\frac{216}{3}-216+216=](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bx%5E3%7D%7B3%7D-6x%5E2%2B36x%5D%5Climits%5E6_0%3D%28%5Cfrac%7B6%5E3%7D%7B3%7D-6%286%29%5E2%2B36%286%29%29-%280%29%3D%5Cfrac%7B216%7D%7B3%7D-216%2B216%3D)
the area under the curve bounded by the lines and the x axis is 72 square units
You just need to try with all of the pair f.ex
0,-2 -> put 0 in the x-> y=8*0-2=-2 so ok it is q possible pair of the function..
You try every pair till you find set where all the pairs fit
So set 1
1 and 6: 8*1-2=6 ok
2 and 14 : 8*2-2=14 ok
Hope I could help
Answer:
The amount of work required to remove all of the water by pumping it over the side = 3.12 × 10⁶ lbs.ft
Step-by-step explanation:
Work done in moving anything from point A to point B = Fx
For this setup,
If we take an elemental vertical height, dx, The volume would be A.dx
where A = Cross sectional Area = πr² = π(18)² = 1017.88 ft²
dV = 1017.88 dx
The elemental force on that part will be
dF = ρg dV
ρg = 63.8 lbs/ft³
dF = 63.8 × 1017.88 dx = 64940.5 dx
F = ∫dF = 64940.5 dx
W = Fx = (∫dF)x = ∫ 64940.5x dx = 64940.5 ∫ xdx
We'll be integrating from (11 - 6) ft to 11 ft because that's the total height it'll be pumped through
W = 64940.5 (x²/2)¹¹₅ = 64940.5((11² - 5²)/2) = 3.12 × 10⁶ lbs.ft
Patrick's statement is incorrect.
Explanation:
Crackers come in package of = 24
Cheese slices come in package of = 18
As given, Andy wants one cheese slice for each cracker so we will find the least common multiple of 24 and 18.
The L.C.M is 72.
So, Andy needs 72 crackers and 72 cheese slices.
Number of cracker packages needed=
Number of cheese slice packages needed=
Hence, Patrick's statement is incorrect. Andy needs 3 packages of crackers and 4 packages of cheese slices to make them even.
