So.....the area of the cross-section [ X Y X' Y'] = 80
This is actually a rectangle that has the dimensions 2r and h
So we have that 2r * h = 80 → rh = 40 → h = 40 / r
The total surface area =
2pi r ( r + h) = 2 pi r ( r + 40/r) = 2 pi r * [ ( r^2 + 40)/r ] = (2pi) ( r^2 + 40) = 2pi r^2 + 80pi
Since 2 pi r^2 is the area of the top and bottom of the cylinder, then 80 pi must be the other part of the surface area = the lateral surface area
Guest is probably correct......I don't know if we can consider single digits as having a sum !!!
Mmmm...let's see ( I'm not sure myself !!!)
The first 9 positive integers are all divisible by their sum
Then 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, 48, 50
I get 23 !!!
a) We have that
60 = k (2 -1)^3 (2 +3) simplify
60 = k ( 1) ^3 * 5 divide both sides by 5 and simplify
60 = 5 k divide both sides by 5
k =12
b ) x intercepts = -3 and 1
y intercepts....let x = 0 and we have that 12 ( -1)^3 (3) = -36
c) Here's the graph : https://www.desmos.com/calculator/aucqv32fyl
Thanks, Melody....and yeah, Cal....I DO exist (but just barely....LOL!!!! )
Call Sarah's score, S
The Greg's score = S - 40
So....we have that [ Sarah's score + Gref's score ] / 2 = 102
[ S + S - 40 ] /2 = 102 multiply both sides by 2
2S - 40 = 204 add 40 ro both sides
2S = 244 divide both sides by 2
S = 122 = Sarah's score
The total surface area is ( 2* pi * r ) ( r + h)
The lateral surface area is (2 * pi * r ) * h = .80 ( 2 * pi * r) ( r + h)
Divide out 2* pi * r from both sides and we have that
h = .80 ( r + h) simplify
h = .80r + .80h
h - .80 h = .80r
.20 h = .80 r divide both sides by .80
(1/4) h = r divide both sides by h
(1/4) = r / h [ the base radius is 1/4 the height ]
I like this one....
If Betty finishes 20m behind Wilma in the original 100m race.....her rate must be (80/100) = 4/5 of Wilma's
Thus, when Wilma runs 1 m , Betty runs (4/5) m
And in the second race, Wilma now runs 120 m
But Betty only runs (4/5) of this = (4/5) (120) = 96 m
So....Betty is 100 - 96 = 4m from the finish line when Wilma crosses it
Midpoint of S1 = [ (3 + sqrt (2) + 4)/2 , ( 5 + 7)/2 ] = [ (7 + sqrt (2)) / 2 , 6 ]
Midpoint of S2 = [ ( 6 - sqrt (2) + 3 ) / 2 , (8 + 15) / 2 ] = [ (9-sqrt (2)) / 2, 23/2 ]
Midpoint of the segment with endpoints at the midpoints of s_1 and s_2
[ (7 + sqrt (2) + 9 - sqrt (2) ) / 2 , ( 6 + 23/2) / 2 ] =
[ 16/2 , (12/2 + 23/2) / 2 ] =
( 8 , 35 / 4 )
This seems trickier than it really is.......
A rational function will have a horizontal asymptote of 0 whenever the degree of the polynomial in the numerator is less than the degree of the ploynomial in the denominator
So....all we need to have g(x) be is something where we change the signs on the 2x^4, -8x^3 and x^2 terms in the original polynomial (keeping the same denominator)
So....let g(x) be
(-2x^4 + 8x^3 - x^2 )
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(x^2 + x - 2)
When we add this to f(x) we get
(7x - 3)
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And this function will have a horizontal asymptote of 0 as x approaches infinity
