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3 years ago

Determine if the set of ordered pairs represents a function. Explain why or why not.

Mathematics

1 answer:

Luden [163]3 years ago

7 0

No it doesn’t because in a function you can’t use a number twice on an input or output

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(-10p+9=12) I need somebody to solve this and show me the work of this equation.

Misha Larkins [42]

-10p+9=12
-9 -9
-10p=3
—— —
-10 -10
P=3/-10 or -0.3

8 0

3 years ago

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Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

Step(ii):-

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

3 0

3 years ago

Write the equation ∣4−r/3∣=7 as two separate equations, and enter each equation in its own answer box below. Neither of your equ

Naddik [55]

Answer:

4 - r/3 = 7 and 4 - r/3 = -7

Step-by-step explanation:

absolute value equations get split into a positive equation and a negative equation

4 - r/3 = 7 and 4 - r/3 = -7

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3 years ago

Relationship A has a greater rate than Relationship B. This table represents Relationship B.

exis [7]

Answer:

y=16.4x I think it's wrong

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3 years ago

6 13 multiply by 2 and then add 1

algol13

Answer:

1227 is correct ................................................

Step-by-step explanation:

6 0

3 years ago

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