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Paul [167]
3 years ago
15

What is the domain and range of relation (4,3)(-2,2)(5,-6)

Mathematics
1 answer:
vazorg [7]3 years ago
5 0
Domain: {-2, 4, 5}
Range {-6, 2, 3}
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Question 5 (Solving Problema) lon 5 Solving Problems] Write 24 18 in slmplest form​
otez555 [7]

Answer:

4/3

Step-by-step explanation:

reduce the fraction by a factor of 6

pretty sure you mean write 24/18 in simplest form so that's the answer

Hope this helps! :)

8 0
2 years ago
Find the value of the integral that converges.<br> ∫^-5_-[infinity] x^-2 dx.
Bingel [31]

Answer:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

Step-by-step explanation:

For this case we want to find the following integral:

\int_{-\infty}^{-5} x^{-2}dx

And we can solve the integral on this way:

\int_{-\infty}^{-5} x^{-2}dx= \frac{x^{-2+1}}{-2+1} \Big|_{-\infty}^{-5}

\int_{-\infty}^{-5} x^{-2}dx= -\frac{1}{x} \Big|_{-\infty}^{-5}

And if we evaluate the integral using the fundamental theorem of calculus we got:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

8 0
3 years ago
Simplify the following expression.<br> 5 [13 + 10 = (3 + 2)] + 9 x 2
Sergio [31]

Answer:

i BELIEVE ITS

Step-by-step explanation:

5 [13 + 10 = (3 + 2)] + 9 x 2

65+50=25+18

115=43

7 0
3 years ago
Suppose f (x) = x^3 . Find the graph of f (x + 1)
ira [324]

Answer:

its 3rd graph, the one with (-1,0) point

3 0
3 years ago
Read 2 more answers
1 point<br> 10. What is the solution of the system of equations graphed below?*
Tema [17]

Answer:

No solution

Step-by-step explanation:

Unfortunately your screenshot has cut off the final answer choice, if it says "no solution" you should choose that because these two lines are parallel. Meaning they will never intersect, and the point of intersection is usually the solution.

5 0
3 years ago
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