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When x = 7, f(x) = 3*(4-x) = 3*(-3) = -9
f(x) = 2 * 2 - x + 9 f(-4) = 2 * 2 -(-4) + 9 f(-4) = 4 + 4 + 9 = 17
Let F = 1.7666.... then 10*F = 17.6666.... So 10*F - F = 17.6666.... - 1.7666... 9*F = 15.9 F = 15.9/9 = 159/90 = 53/30
3 f'(1) is the slope of the tangent line of the curve at x=1. Because the tangent line at x=1 goes through (1,7) and (-2,-2), its slope is (7-(-2))/(1-(-2))=9/3=3. So f'(1)=3.
By using the fundamental theorem of Calculus. i.e. The integral of f(x) = F(x), your limits are [a,b]. Solve: F(b) - F(a). The FTC, second part, says that if f is a continuous real valued function of [a,b] then the integral from a to b of f(x)= F(b) - F(a) where F is any antiderivative of f, that is, a function such that F'(x) = f(x). Example: Evaluate the integral form -2 to 3 of x^2. The integral form -2 to 3 of x^2 = F(-2) - F(3) = -2^3/3 - 3^3/3 = -8/3 - 27/3 = -35/3
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The answer is -1/3. Justofication: Suppose F = -0.333... then 10*F = -3.333... Subtracting the first from the second: 9*F = -3 and so F = -3/9 = -1/3
3f+9 = 23 3f = 23-9 3f = 14 f = 14/3 3f+9=23 -9 -9 3f= 14 /3 /3 f= 4.66666 (6 repeats)
Suppose f = 0.33... recurring then 10*f = 3.33... recurring Subtracting the first equation from the second, 9*f = 3 so f = 3/9 = 1/3.
To find f(-3), we substitute -3 into the function f(x) = x^2 + x: f(-3) = (-3)^2 + (-3) = 9 - 3 = 6
12f -9 = 27 12f = 27+9 12f = 36 f = 3
f(x) = 3x + 9 y = 3x + 9 so 3x = y - 9 x = y/3 - 3 Therefore, the inverse function is g(x) = x/3 - 3
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When x = 7, f(x) = 3*(4-x) = 3*(-3) = -9
The GCF is 3.
16.66 C = (16.66 x 9/5) + 32 F = (50/3 x 9/5) + 32 F = 62 F
kq1q2/d = F 10^-12(k)/d = F 10^-12(9*10^9)/d = F 9*10^-3 / 3*10^-2 = F 3*10^-1N = F