If you wanted to rotate f(x) around the y axis from x=a to x=b then the volume would be
that would be
1/x^10=x^-10
integrate
-1/(9x^9)
so
=
![\pi [\frac{-1}{9x^{9}}]^8_1=\pi (\frac{-1}{9(8^{9})}-\frac{-1}{9(1^{9})})=](https://tex.z-dn.net/?f=%5Cpi%20%5B%5Cfrac%7B-1%7D%7B9x%5E%7B9%7D%7D%5D%5E8_1%3D%5Cpi%20%28%5Cfrac%7B-1%7D%7B9%288%5E%7B9%7D%29%7D-%5Cfrac%7B-1%7D%7B9%281%5E%7B9%7D%29%7D%29%3D)
that's the volume of the solid
When x gets large, you can ignore everything except the highest-degree terms of the numerator and denominator. Effectively, you want the value of
-2x/x
= -2
The horizontal asymptote is -2.
The minute hand moves more than 24 inches in one hour.
Answer:
FH = 108
Step-by-step explanation:
The given figure requires we use the Pythagorean theorem to write two relations involving right triangle side lengths. The Pythagorean theorem tells us the square of the hypotenuse is the sum of the squares of the other two sides.
__
<h3>Triangle EGH:</h3>
EG² = GH² +HE²
GH² = EG² -HE² = 53² -28² = 2025 . . . . . solve for GH², use given values
<h3>Triangle FGH:</h3>
FG² = GH² +FH²
FH² = FG² -GH² = 117² -2025 = 11664 . . . . solve for FH², use known values
FH = √11664 = 108 . . . . . take the square root
The length FH is 108.
Answer:
D) 50°. You got parallel lines cut by a traversal, so angles 4 and 8 are congruent because they're corresponding angles.
Step-by-step explanation:
