Plug in the known variables.
5n + 2 n = number of games bowled
5(3) + 2 = 15 + 2 = 17
(a) See the attached sketch. Each shell will have a radius <em>y</em> chosen from the interval [2, 4], a height of <em>x</em> = 2/<em>y</em>, and thickness ∆<em>y</em>. For infinitely many shells, we have ∆<em>y</em> converging to 0, and each super-thin shell contributes an infinitesimal volume of
2<em>π</em> (radius)² (height) = 4<em>πy</em>
Then the volume of the solid is obtained by integrating over [2, 4]:
(b) See the other attached sketch. (The text is a bit cluttered, but hopefully you'll understand what is drawn.) Each shell has a radius 9 - <em>x</em> (this is the distance between a given <em>x</em> value in the orange shaded region to the axis of revolution) and a height of 8 - <em>x</em> ³ (and this is the distance between the line <em>y</em> = 8 and the curve <em>y</em> = <em>x</em> ³). Then each shell has a volume of
2<em>π</em> (9 - <em>x</em>)² (8 - <em>x</em> ³) = 2<em>π</em> (648 - 144<em>x</em> + 8<em>x</em> ² - 81<em>x</em> ³ + 18<em>x</em> ⁴ - <em>x</em> ⁵)
so that the overall volume of the solid would be
I leave the details of integrating to you.
We have that
<span>points A (-5, 6) and B (7, -1)
Part A)
find the distance
d=</span>√[(y2-y1)²+(x2-x1)²]-------> d=√[(-1-6)²+(7+5)²]----> d=√(49+144)
d=√193 units
Part B)
find the midpoint
ABx=(x1+x2)/2-----> (-5+7)/2-----> ABx=1
ABy=(y1+y2)/2-----> (-1+6)/2-----> ABy=2.5
the midpoint is (1,2.5)
Part C)
find the slope
m=(y2-y1)/(x2-x1)-----> m=(-1-6)/(7+5)--------> m=-7/12
the slope m=-7/12
A. The discriminant is 81. The formula is b^2 - 4ac.
B. 2 answer and both will be rational due to the fact that the discriminant is a perfect square.
C. Solutions are 1/2 and -4. You can find using the quadratic formula.
Answer:
B
Step-by-step explanation:
We know that 3² = 9, so for g(x) to pass through (3,3), we need to multiply x² by 1/3, giving option B.