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vova2212 [387]
3 years ago
10

If f(n) = n2 - 2n, which of the following options are correct? Select all that apply. f(2) = 0 f(1) = 3 f(-4) = 24 f(5) = 35 f(-

2) = 0
Mathematics
2 answers:
iren [92.7K]3 years ago
5 0

Answer:

f(2)=0

f(-4)=24

Step-by-step explanation:

We are given

f(n)=n^2-2n

We are asked which of the following are true?

f(2)=0

f(1)=3

f(-4)=24

f(5)=35

f(-2)=0

Let's start with f(2)=0.

f(2)=2^2-2(2)

f(2)=4-4

f(2)=0

So f(2)=0 is true.

Moving on to f(1)=3.

f(1)=1^2-2(1)

f(1)=1-2

f(1)=-1

So f(1)=3 is false.

Now for f(-4)=24.

f(-4)=(-4)^2-2(-4)

f(-4)=16+8

f(-4)=24

So f(-4)=24 is true.

f(5)=35?

f(5)=5^2-2(5)

f(5)=25-10

f(5)=15

So f(5)=35 is false.

Finally f(-2)=0.

f(-2)=(-2)^2-2(-2)

f(-2)=4+4

f(-2)=8

So f(-2)=0 is false.

xz_007 [3.2K]3 years ago
3 0

Answer:

The correct option are,

f(2) = 0, and f(-4) = 24

Step-by-step explanation:

Given function,

f(n) = n^2 - 2n

By substituting n = 2,

f(2) = 2^2 - 2(2) = 4 - 4 = 0

By substituting n = 1,

f(1) = 1^2 - 2(1) = 1 - 2 = -1

By substituting n = -4,

f(-4) = (-4)^2 - 2(-4) = 16 + 8 = 24

By substituting n = 5,

f(5) = (5)^2 - 2(5) = 25 - 10 = 15

By substituting n = -2,

f(-2) = (-2)^2 - 2(-2) = 4 + 4= 8

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6 0
3 years ago
8/x-5 - 9/x-4 = 5/x^2-9x+20
adelina 88 [10]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2752942

_______________


Solve the equation:

\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}


Reduce the fractions at the left side so that they have the same denominator:

\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}


Numerators must be equal:

\mathsf{8x-32-(9x-45)=5}\\\\
\mathsf{8x-32-9x+45=5}\\\\
\mathsf{8x-9x=5+32-45}\\\\
\mathsf{-x=-8}\\\\
\mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}


I hope this helps. =)


Tags:  <em>rational equation fraction solution algebra</em>

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Read 2 more answers
Does anyone know how to answer these two questions?
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