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 log_{7} 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

subtract them:

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
thank you!

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