Select Is a Function or Is not a Function to correctly classify each relation.
<span><span>Title Is a Function Is not a Function</span><span><span><span><span>{<span><span>(<span>3, 7</span>)</span>,<span>(<span>3, 6</span>)</span>,<span>(<span>5, 4</span>)</span>,<span>(<span>4, 7</span>)</span></span>}</span></span>
</span><span><span><span>{<span><span>(<span>1, 5</span>)</span>,<span>(<span>3, 5</span>)</span>,<span>(<span>4, 6</span>)</span>,<span>(<span>6, 4</span>)</span></span>}</span></span>
</span><span><span><span>{<span><span>(<span>2, 3</span>)</span>,<span>(<span>4, 2</span>)</span>,<span>(<span>4, 6</span>)</span>,<span>(<span>5, 8</span>)</span></span>}</span></span>
</span><span><span><span>{<span><span>(<span>0, 4</span>)</span>,<span>(<span>3, 2</span>)</span>,<span>(<span>4, 2</span>)</span>,<span>(<span>6, 5</span>)</span></span>}</span></span>
</span></span></span>
Hello!

First you had to used apply rule -(-b)=b

Since the denominators are equal to it should be combine by the fractions.

Then add 4+2=6


Finally you had to cancel by the common factor of 3.

Hope this helps! And thank you for posting at here on Brainly. And have a great day! -Charlie :)
Answer:
The function a (t) is a vector function composed of the component functions
and
. How
are infinitely derivable functions in R, so they are regular functions in R.
Now, for
, you have to
. How the functions
are periodic functions with period
the vector function
will take the same point
at
then the vector function is auto-intercepted
Step-by-step explanation:
Answer:
its either A/B
Step-by-step explanation: