Posted by ANR on Sunday, October 14, 2012 at 10:16pm.
1)let k and w be two consecutive integers such that k<x<w. If log base 7 of 143 = x, find the value of k+w
2) if 7 and -1 are two of the solutiosn for x in the equation 2x^3 +kx^2 -44x+w=0, find the value of k+w
3) from an ordinary deck of 52 cards, two cards are selected at random (without replacement). find the probability that both cards were hearts. express your answer as a common fraction reduced to lowest terms.
- Math - Reiny, Sunday, October 14, 2012 at 10:35pm
1. if log7 143 = x
then 7^x = 143
but 7^2 = 49 and 7^3 = 343
so k=2 and w = 3
then k+w = 5
2. let f(x) = 2x^3 + kx^2 - 44x + w
if 7 is a solution then f(7) = 0
2(343) + 49k - 308 + w = 0
49k + w = -378
if -1 is a solution, then f(-1) = 0
2(-1) + k + 44 + w = 0
k + w = -42
48k = -336
k = -7
in k+w=-42, -7+w = -42 ---> w = -35
and k+w = -42
3) prob(2 hearts) = (13/52)(12/51) = 1/17
- Math - ANR, Monday, October 15, 2012 at 12:04am
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