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kirill [66]
2 years ago
9

PLEASE HELP IS THIS CORRECT GUYS YES OR NO!

Mathematics
1 answer:
mote1985 [20]2 years ago
4 0

Answer:

yes

Step-by-step explanation:

i think if it's wrong pls don't hate me

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Which system of linear inequalities is graphed?
ANEK [815]

To determine the system of inequality that has been graphed, we need to be very smart.


Observe that all the equations are the same.


Also note that all the lines are solid. This should tell you that the inequities should involve.


\ge \:or\: \le



Therefore the answer is between option A and B.



So we choose a point in the solution region to discriminate between A and B.


Let us choose the origin since that is easy to evaluate.


We substitute into option A to get.

y\le 3x-1\:and\: x+ 3y \ge 6




0\le 3(0)-1\:and\: 0+ 3(0) \ge 6




0\le -1\:and\: 0\ge 6


The above statements are false



We now substitute into option B



y\ge 3x-1\:and\: x+ 3y \le 6




0\ge 3(0)-1\:and\: 0+ 3(0) \le 6




0\ge-1\:and\: 0 \le 6


Both statements are true, therefore the correct answer is option B.




8 0
3 years ago
If f(x)=1/9x-2 what is f^-1(x)
Katarina [22]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
Solve for y.
DochEvi [55]
The answer is b y=6 Use photonath if any more equations like this one
5 0
3 years ago
Read 2 more answers
Find the inverse: f(x)=x+5/3x-1<br><br> f^-1x=
babunello [35]

We have f(x)=\dfrac{x+5}{3x-1}.

To find inverse function f^{-1}(x) we substitute x with f^{-1}(x) and vice-versa to get

x=\dfrac{f^{-1}(x)+5}{3f^{-1}(x)-1}

Now solve for f^{-1}(x). Note that I will use j instead.

x=\dfrac{j+5}{3j-1} \\x(3j-1)=j+5 \\3jx-x=j+5 \\3jx-x-j-5=0 \\3jx-j=x+5 \\j(3x-1)=x+5 \\j=\dfrac{x+5}{3x-1}

So we find that f(x)=f^{-1}.

Hope this helps.

7 0
3 years ago
Read 2 more answers
Given that f(a+b)= f(a) + f(b) and f(x) is always positive, what os the value of f(0)?
FrozenT [24]
This is a strange question, and f(x) may not even exist. Why do I say that? Well..

[1] We know that f(a+b) = f(a) + f(b). Therefore, f(0+0) = f(0) + f(0). In other words, f(0) = f(0) + f(0). Subtracting, we see, f(0) - f(0) = f(0) or 0 = f(0).

[2] So, what's the problem? We found the answer, f(0) = 0, right? Maybe, but the second rule says that f(x) is always positive. However, f(0) = 0 is not positive!

Since there is a contradiction, we must either conclude that the single value f(0) does not exist, or that the entire function f(x) does not exist.

To fix this, we could instead say that "f(x) is always nonnegative" and then we would be safe.
4 0
3 years ago
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