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olya-2409 [2.1K]
3 years ago
12

Let f(x)=5x+2 and gx=x^2-4x+1

rac{f}{g}(5)" alt="\frac{f}{g}(5)" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Rasek [7]3 years ago
3 0

Answer:

find f(5) and g(5) then divide f(5) by g(5) i.e f(5)/g(5) Answer will be 25/6

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Harry got 42 out of 49 correct in his test.
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42/49 is the fraction

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Round 32.697 to the nearest tenth hundredth and whole number
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3 years ago
Given f(x)=8x^5, find f^-1(x) <br> Then state whether f^1(x) is a function.
irina1246 [14]

Answer:

f^-1(x) = (x/8)^(1/5) is a function

Step-by-step explanation:

The inverse of a function is the function reflected across the line y=x. This results in the coordinate points (x,y) of the function becoming (y,x) for the inverse function. Algebraically to find the inverse, switch the x and y locations and solve for y.

y = 8x^5

x = 8y^5

x/8 = y^5

(5)√(x/8) = y

This is a fifth root of (x/7) also written in exponents as y = (x/8)^(1/5).

This is a function. While it appears not to be a function because the middle portion over the origin appears vertical, it is a function because the middle portion over the origin is changing and graphing software shows it has no input with more than one output. Without graphing software you would rule it is not a function.

6 0
3 years ago
What is the vertex of the quadratic function f(x)=(x-8)(x-2)
Mrac [35]

So firstly, foil (x-8)(x-2): f(x)=x^2-2x-8x+16\\ f(x)=x^2-10x+16

Next, remember that the vertex lies on the axis of symmetry, and the equation to find the axis of symmetry is x=\frac{-b}{2a} (a = x^2 coefficient, b = x coefficient). Solve for the axis of symmetry as such:

x=\frac{10}{2*1}=\frac{10}{2}=5

So now that we know that the axis of symmetry is x = 5, we also know that's the vertex's x-coordinate (since the vertex falls on the axis of symmetry). Now plug in 5 for x in the original equation to solve for f(5):

f(5)=5^2-10(5)+16\\ f(5)=25-50+16\\ f(5)=-9

So putting it together, the vertex of this equation is (5,-9).

4 0
3 years ago
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