7 is infinite 6 is 4/3 5 is zero 4 is negative 5/4
Substitute y=4x to the second equation:
x^2 + (4x)^2 = 17
x^2 + 16x^2 = 17
17x^2 = 17
x^2 = 17/17
x^2 = 1
x = 1 and -1
When x=1, y=4(1) = 4
When x=-1, y=4(-1) = -4
Thus the solutions would be (1,4) and (-1,-4). That would correspond to D. and A.
Answer:
its the last one
Step-by-step explanation:
Answer:
A = [- 2Bl + U^2 +/- √((2BL - U^2)^2 - 4B^2L^2) ] / 2L
Step-by-step explanation:
L = U^2A / (A + B)^2
U^2 A = L(A^2 + 2AB + B^2)
U^2 A = LA^2 + 2ABL + B^2L
LA^2 + 2ABL - U^2A + B^2 L = 0
LA^2 + (2BL - U^2)A + B^2L = 0
Using the quadratic formula
A = [- (2BL - U^2) +/- √[(2BL - U^2)^2 - 4*L*B^2L] / 2L
A = [- 2BL + U^2 +/- √[(2BL - U^2)^2 - 4B^2L^2) ] / 2L
