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3 years ago

Which set of odered pairs is not a function?

Mathematics

2 answers:

harkovskaia [24]3 years ago

8 0

Answer:

your answer is A, im sure of

Step-by-step explanation:

andriy [413]3 years ago

4 0

Answer:

Set A (Top Left)

Step-by-step explanation:

A function's inputs is assigned to exactly one output. This means a function's inputs must not repeat.

The input of '3' repeats in the top left set given, so it is not a function.

{(<u>3</u>,1), (2,1), (1,2), (<u>3</u>,2)}

Hope this helps you.

You might be interested in

Solve for x.

astra-53 [7]

Answer:

Option: C is correct.

$x=\ln\sqrt{\dfrac{t+1}{1-t}}$

Step-by-step explanation:

We are asked to solve for 'x' such that we are given a equation as:

$\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=t$

This equation could also be written as:

$e^x-e^{-x}=t(e^x+e^{-x})\\\\e^x-e^{-x}=te^x+te^{-x}\\\\e^x-te^x=te^{-x}+e^{-x}\\\\(1-t)e^x=(t+1)e^{-x}\\\\e^{2x}=\dfrac{t+1}{1-t}$

on taking logarithmic function on both the sides we have:

$2x=\ln (\dfrac{t+1}{1-t})$

(since $\ln (e^x)=x$ )

$x=\dfrac{1}{2}\times \ln (\dfrac{t+1}{1-t})$

as we know $\dfrac{1}{2}\times \ln b = \ln (b^\frac{1}{2})$

Hence, we have: $x=\ln\sqrt{\dfrac{t+1}{1-t}}$

Hence, option C is correct.

5 0

3 years ago

Read 2 more answers

Find the area of the following regions, expressing your results in terms of the positive integer n â‰¥ 2. The region bounded by

cluponka [151]

Answer:

The area of the searched region is $A= a+b+ \frac{2n}{n+1}- \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1}-2$

Step-by-step explanation:

If you want to find the area of a region bounded by functions f(x) and G(x) between two limits (a,b), you have to do a double integral. you must first know which of the functions is greater than the other for the entire domain.

In this case, for 0<x<1, f(x)<g(x)

while for 1<x, g(x)<f(x).

Therefore if our domain is all real numbers superior to 0 (where the limit 0<a<1 and 1<b), we have to do 2 integrals:

A=A(a<x<1)+A(1<x<b)

$A(a$

$A(1$

$A=a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1} + b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1} = a+b+ \frac{2n}{n+1} - \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1} -2$