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dexar [7]
2 years ago
7

If g(x) =f(x+k) , what is the value of k.

Mathematics
2 answers:
Umnica [9.8K]2 years ago
4 0

Answer:

k = 6

Step-by-step explanation:

g(x) = f(x+6)

JulijaS [17]2 years ago
3 0

Answer:

k = 6

Step-by-step explanation:

g(x) = f(x+6)

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What is the earned income credit (EIC)?
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Answer:

D

Step-by-step explanation:

The United States federal earned income tax credit or earned income credit is a refundable tax credit for low- to moderate-income working individuals and couples, particularly those with children.

7 0
3 years ago
What is the value of a in the equation −14a−5=−12?
-BARSIC- [3]
-14a-5=-12
+5 +5
-14a=-7
Divide -7 by -14a
a=2
7 0
2 years ago
How to algebraically find the limit of a function as x approaches infinity?
Sauron [17]

Well... One way you can do this is by testing a set of arrays and see the trend. If I chose to find what y1 is in (100, y1) and what y2 is in (101, y2), I would find the difference between y2 and y1. If y2 - y1 is positive, this means there is a positive relationship and y is also approaching POSITIVE infinity. A negative relation means that it is approaching NEGATIVE infinity. However, it could be approaching a single number like "4" for instance, and you just need to plug in the right number of data sets to make that educated guess.

Formula Example:
5 + 1 / (x + 1) will always approach 5 because "1 / (x + 1) will approach 0".

Hope this helps.

5 0
3 years ago
EMERGENCY PLEASE HELP! Which set of ordered pairs does not represent a function?
lilavasa [31]

Answer:

c.

Step-by-step explanation:

The X value 1 is repeated

7 0
2 years ago
Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra
Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize x^2-\frac13 as yet another difference of squares:

x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)

And if you're working over the field of complex numbers, you can go even further. For instance,

x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)

But I think you'd be fine stopping at the first result,

x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}

6 0
3 years ago
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