y = f(x) = (x-5)2. At x = 5, the y value is zero, so the point (5,0).
y - x5 is an expression, not an equation. Furthermore, it is polynomial of order 5 and a non-linear function cannot be expressed in slope-intercept form.
1,565 = x5 1,5651/5 = (x5)1/5 1,5651/5 = x5/5 1,5651/5 = x 4.3541 = x
11x5=55 !
Assuming that the question is find x in x5=1610; the answer is 4.378901
1
y - x5 is an expression, not an equation. Furthermore, it is polynomial of order 5 and a non-linear function cannot be expressed in slope-intercept form.
1,565 = x5 1,5651/5 = (x5)1/5 1,5651/5 = x5/5 1,5651/5 = x 4.3541 = x
11x5=55 !
Assuming that the question is find x in x5=1610; the answer is 4.378901
x10 = x5.x5 = (x5)2 [x power five whole squared] Equation is x10 + x5 - 2 Replacing x10 (x5)2+x5 - 2 Substituting x5 by Y Equation becomes Y2 + Y - 2 = 0 Y2 + 2Y - Y - 2 = 0 Y(Y + 2) - 1(Y+2) = 0 (Y-1)(Y+2) = 0 The values of Y are 1 and -2 Y = x5 = 1 Therefore, x = 1 Y = x5 = -2 x = fifth root of -2, which is an imaginary value.
512 x 5 = 2,560
1
5 times 5 equals 25.
a(x5) + b(x5) + c(x5) + d(x5) + e(x5) = abcde(a+b+c+d+e) x5 = abcdeThis equation has at least 5 variables. To solve for all of them requires at least 4 more equations.
-1
x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4
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