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07282007  #1 
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i need algebra help!!!?
problem 58 says: find the equation of a line that is perpendicular to y=1/5x+11 but goes trough the piont (6,8)#61: a) x=? equationfx)=3x7 f(x)=1 b) x=? equation: 2x6 (2x6 is in the square root sign) f(x)=10(you have to find what x=to56: simplify the rational expression. what valuses can x not be?3xsquare+11x20 (over) 2xsquare+11x+5

07282007  #2 
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58. to get the slope of the perpendicular line you take the reciprocal of 1/5 and change the sign. so the slope is 5. then use the equation (yy1) = m(xx1) where m=5, y1=8, and x1=6. so thats (y8) = 5(x  6) + 11 which is the same as y = 5x + 3361. just think of f(x) as y soy=3x7 and y=1plug in 1 to the first equation1=3x7solve for x6=3xx=256. that factors out to (3x4)(x+5) / (2x+1)(x+5)the factors (x+5) cancel out so thats (3x4) / (2x+1)the bottom part, (2x+1), cant equal 0so 2x+1=0 when x = 1/2

07282007  #3 
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I'm not too sure what you trying to ask in 61 a & bbut I can help you with #56(3x2+11x20 ) / (2x2+11x+5)= (3x2+ 15x â€“ 4x 20 ) / (2x2+10x+ 1x +5)= [3x(x+5) â€“ 4(x+5) ] / [ 2x (x+5) + 1 (x+5)]= (3x  4) /(2x + 1)Setting both sides equal to zero and then simplifying it more= 3x 4 = 0 and 2x +1 = 0= 3x = 4 and 2x = 1= x 4/3 and x = 1/2 Setting X = to these values will give us a zero in the denominator and the numerator. And hence X cannot be equal to 4/3 and 1/2 Hope this helps

07282007  #4 
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58.)y = (1/5)x + 11(6,8), m = 58 = 5(6) + b8 = 30 + b22 + bANS : y = 5x  2261.)a.)f(x) = 3x  7f(x) = 11 = 3x  76 = 3xx = 2b.)f(x) = sqrt(2x  6)f(x) = 1010 = sqrt(2x  6)2x  6 = 1002x = 106x = 5356.)(3x^2 + 11x  20)/(2x^2 + 11x + 5)((3x  4)(x + 5))/((2x + 1)(x + 5))(3x  4)/(2x + 1)x cannot be 5 or (1/2)

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