Find all values of b such that the minimum distance from the point (2,0) to the line Y = (4/3)X + b is 5.
d = 5 = ((Y - 0)[sup:u4ekfm1r]2[/sup:u4ekfm1r] + (X - 2)[sup:u4ekfm1r]2[/sup:u4ekfm1r])[sup:u4ekfm1r]1/2[/sup:u4ekfm1r]
Squaring both sides,
25 = (Y - 0)[sup:u4ekfm1r]2[/sup:u4ekfm1r] +...
Let f(n) = n[sup:2bpiakor]2[/sup:2bpiakor] - n + 2
Find a value for k such that the equation
f(n[sup:2bpiakor]2[/sup:2bpiakor] + k) = f(n) X f(n + 1)
holds for all values of n.
f(n[sup:2bpiakor]2[/sup:2bpiakor] + k) = (n[sup:2bpiakor]2[/sup:2bpiakor] + k)[sup:2bpiakor]2[/sup:2bpiakor] -...
Let f(x) = 2x + 3
Find values for a and b such that the equation f(ax + b) = x is true for all values of x.
2(ax + b) + 3 = x
So, how does one solve for three variables without some method of substitution?
Or, have I missed the substitution possibilities?
Find a point P on the curve y = x[sup:1o6xq0tm]3[/sup:1o6xq0tm] such that the slope of the line passing through P and (1, 1) is 3/4.
p[sub:1o6xq0tm]1[/sub:1o6xq0tm] = (1, 1)
p[sub:1o6xq0tm]2[/sub:1o6xq0tm] = (x, x[sup:1o6xq0tm]3[/sup:1o6xq0tm])
m = (y[sub:1o6xq0tm]2[/sub:1o6xq0tm] -...
X^5 - 2X^3 - 2X = 0
I factored X out of the left side,
X(X^4 - 2X^2 - 2) = 0
So one solution is X = 0, which obviously checks in the original equation.
That leaves
X^4 - 2X^2 - 2 = 0,
And I have no idea what to do next
Factor 8 - (a + 1)^3
I started cubing the (a + 1) binomial, resulting in
8 - (a^3 + 3a^2 + 3a + 1), and then subtracting to get
8 - a^3 - 3a^2 - 3a - 1, and then subtracting 1 from 8 I get,
7 - a^3 - 3a^2 -3a
I do not see what to do next.
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