Question 68140
For f(x)=1/x^2-2x-8, find the interval(s) where f(x)<0
The fraction is negative when the denominator is negative.
The denominator is negative when (x-4)(x+2)<0
Draw a number line and mark -2 and +4
Check a test value in each interval to see where the solution set is:
If x=-100 (x-4)(x+2)is positive so no solutions in (-inf,-2)
If x=0, (x-4)(x+2) is negative so (-2,4) is part of the solution set.
If x=100, (x-4)(x+2) is positive so no solutions in (4,inf(
Final: f(x)is negative when x in in the interval (-2,4)
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Express the following statement as a formula with the value of the constant of proportionality determined with the given conditions: w varies directly as x and inversely as the square of y. If x=15 and y=5, then w=36.
w=kx/y^2 ; 36=k*15/25 = k=60
Therefore w=(60x/y^2)
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The electrical resistance R of a wire varies directly as its length L and inversely as the square of its diameter. A wire 20 meters long and 0.6 centimeters in diameter made from a certain alloy has a resistance of 36 ohms. What is the resistance of a piece of wire 60 meters long and 1.2 centimeters in diameter made from the same material? 
R=kL/d^2 ; 36 = k*20 m/0.6^2 cm^2; 36=20k/0.36; k=0.648
EQUATION: R =0.648L/d^2
If L=60 and d=1.2, R=0.648*60/(1.2)^2 ; R= 27
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The period of a simple pendulum is directly proportional to the square root of its length. If a pendulum has a length of 6 feet and a period of 2 seconds, to what length should it be shortened to achieve a 1 swcond period? 
Period =k sqrtL
2=ksqrt6
k=2/sqrt6 
k=2sqrt6/6 =(1/3)sqrt6
EQUATIOn:
Period = [(1/3)sqrt6]sqrtL
If Period = 1, 1=[(1/3)sqrt6]sqrtL ; 3/sqrt6 = sqrtL; 9/6=L; L=3/2 ft.
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Cheers,
Stan H.