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lisabon 2012 [21]
3 years ago
15

Can someone help please

Mathematics
1 answer:
devlian [24]3 years ago
3 0

the answer is x=47 hope this helps

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Please help! <br><br> If f (x) = 1/9 x-2 what is f^-1(x)
saw5 [17]

Answer:

Its C or the 3rd option

(9x+2)

Step-by-step explanation:

You have to inverse the equation meaning do everything the opposite so for example, add the 2 to the other side of the equation

4 0
3 years ago
I Really need help with this!
Scrat [10]
Where the heck did x come from!
5 0
3 years ago
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Solve the equation 0.5p -3.45= -1.2.​
ddd [48]

Answer:

4.5

Step-by-step explanation:

0.5p-3.45=-1.2

add 3.45 to -1.2 which would leave the equation to look like

0.5p=2.25

then divide 0.5 to both sides cancelling 0.5 leaving the answer to be

p=4.5

6 0
3 years ago
Does the relation y=x^2-1 defines y as a function of x
Reil [10]

Answer:

Yes.

From Definition - A special relationship where each input has a single output.

Check if there is going to be one y for one x.

Step-by-step explanation:

  1. Re-write it in terms of y=…., meaning solve for y
  2. If it looks like you have two equations then it is a relationship -case a
  3. If you need to prove that one has a function for case b, you would point out that if you substitute any value you will get some value in Reals.
7 0
3 years ago
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If f(x)=2−x12 and g(x)=x2−9, what is the domain of g(x)÷f(x)?
Keith_Richards [23]
\bf \begin{cases}&#10;f(x)=2-x^{12}\\&#10;g(x)=x^2-9\\&#10;g(x)\div f(x)=\frac{g(x)}{f(x)}&#10;\end{cases}\implies \cfrac{x^2-9}{2-x^{12}}

now, for a rational expression, the domain, or "values that x can safely take", applies to the denominator NOT becoming 0, because if the denominator is 0, then the rational turns to undefined.

now, what value of "x" makes this denominator turn to 0, let's check by setting it to 0 then.

\bf 2-x^{12}=0\implies 2=x^{12}\implies \pm\sqrt[12]{2}=x\\\\&#10;-------------------------------\\\\&#10;\cfrac{x^2-9}{2-x^{12}}\qquad \boxed{x=\pm \sqrt[12]{2}}\qquad \cfrac{x^2-9}{2-(\pm\sqrt[12]{2})^{12}}\implies \cfrac{x^2-9}{2-\boxed{2}}\implies \stackrel{und efined}{\cfrac{x^2-9}{0}}

so, the domain is all real numbers EXCEPT that one.
4 0
3 years ago
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