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)=3x-7 f(x)=-1 b) x=? equation: 2x-6 (2x-6 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+11x-20 (over) 2xsquare+11x+5

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 (y-y1) = m(x-x1) where m=5, y1=8, and x1=6. so thats (y-8) = 5(x - 6) + 11 which is the same as y = 5x + 3361. just think of f(x) as y soy=3x-7 and y=-1plug in -1 to the first equation-1=3x-7solve for x6=3xx=256. that factors out to (3x-4)(x+5) / (2x+1)(x+5)the factors (x+5) cancel out so thats (3x-4) / (2x+1)the bottom part, (2x+1), cant equal 0so 2x+1=0 when x = -1/2

I'm not too sure what you trying to ask in 61 a & bbut I can help you with #56(3x2+11x-20 ) / (2x2+11x+5)= (3x2+ 15x – 4x -20 ) / (2x2+10x+ 1x +5)= [3x(x+5) – 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

58.)y = (-1/5)x + 11(6,8), m = 58 = 5(6) + b8 = 30 + b-22 + bANS : y = 5x - 22-----------------------------------------61.)a.)f(x) = 3x - 7f(x) = -1-1 = 3x - 76 = 3xx = 2b.)f(x) = sqrt(2x - 6)f(x) = 1010 = sqrt(2x - 6)2x - 6 = 1002x = 106x = 53-----------------------------------------------56.)(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)