19x^2+3x+2=0 has roots (-3±√143 i)/38 since both roots are complex, if the two quadratics share one root, they share both. So, Ax^2+BX+7 must be a multiple of 19x^2+3x+2. So, it must be 7/2 times, making it 19(7/2)x^2 + 3(7/2)x + 2(7/2) = 66.5x^2 + 10.5x + 7 A+B = 77
you are exactly right tan(4x) = 2tan2x/(1-tan^2(2x)) so, cot(4x) is the reciprocal, as needed.
just apply the double-angle formula for tan(2x) and take the reciprocal
BTW, I was a math major in college. Your original question gave no indication you wanted something in expanded form. If you want a good answer, you need to ask a good question. :-)
The answer (as you know) is $675.505 That would be 600 + 70 + 5 + 0.5 + 0.005
so what did you get? You must not have trusted your calculator to show up here asking for help.
too dumb to use a calculator?
Dang. A typo in the only place it mattered!
as you know, the complex roots come in conjugate pairs. So, the other root is 7-4i f(x) = (x-(7+4i))(x-(7-4i))(x-k) = (x^2-14x+65)(x-k) f(0) = 65(-k) = -2145 so, k = 33 f(x) = (x^2-14x+65)(x-33) = x^3 - 47x^2 + 6527x - 2145
as you know, the slope of y=mx+b is a constant, m. It is the ratio of how fast y changes for every change of 1 in x. So, the slope of this line is (-1-2)/(-4-8) = -3/-12 = 1/4 So, the change in y is 1/4 the change in x. Starting at (8,2), then y-2 = 1/4 (x-8)
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