Question

PLEASE HELP MEEE I HAVE TO GET THESE RIGHT OR ILL FAIL MATH THIS IS MY LAST CHANCE!!!


Select the appropriate restrictions for the variables.


4a+13/a+3



A.


a≠–3


B.


a≠3


C.


a≠4


D.


There are no restrictions on the values of variable a.

Answer 1

woah. hold on let me think i know this

Answer 2

@johnweldon1993 please help me!!!

Answer 3

hmmmmm i dont know sorry :(

Answer 4

Alright...so

You have a fraction...The only thing you need to watch for with a fraction...is the denominator equaling out to 0

Why? You cannot divide a number by 0...it is undefined

So what value for 'a' makes your denominator = 0?

Answer 5

im confused

Answer 6

That's okay..

so we have

\[\large \frac{4a + 13}{a + 3}\] right?

Answer 7

yes

Answer 8

Alright..

Let me put a visual with this...

how does the graph of 1/x look?



Like that

We can see from the graph...when x is any number...we an go up to the function line....but what about when x = 0? what happens we we go to 0 on the x-axis and try to go up or down to meet the function? We cannot do it...the function blows up and become undefined

Does that make sense?

Answer 9

no not at all

Answer 10

omg ima cry im soo confused

Answer 11

Don't cry, We'll get you through it :)

Answer 12

Alright well basically like I've been saying....a denominator of a fraction can NEVER = 0

Because then you get a number divided by 0 ....which is undefined...we do not know it

So with your question

\[\large \frac{4a + 13}{\color \red{a + 3}}\]

\(\large a + 3\) is your denominator....and since we know THAT can NOT equal 0...we do

\[\large a + 3 = 0\]

and now solve for 'a'

What number...can you add to 3...to make 0?

Answer 13

-3

Answer 14

Right....a can NOT = -3 because that will make your denominator = 0...and make this function undefined....

\(\large a \cancel{=} -3\)

Answer 15

omg yay so i do that for a question like


5x-3x^2/2x^4

Answer 16

right?

Answer 17

Is that...

\[\large \frac{5x - 3x^2}{2x^4}\]

right?

Answer 18

yes

Answer 19

Then yes exactly like that...

What will make your denominator = 0?

\[\large 2x^4 = 0\]

What value of 'x' will make that true?

Answer 20

-2x^4?

Answer 21

Not quite...remember this is multiplication hun :) it is 2 times x^4

\[\large 2 \times x^4 = 0 \]

So 2 times what? = 0 ?

Answer 22

2 times 0 =0

Answer 23

Right....so here

\[\large x \cancel{=} 0\]

Answer 24

okay..???

Answer 25

Alright, since your confused, let me ask

is that question the same as this one? "state the restrictions on the variable 'x'?

Answer 26

what does that mean omg lol

Answer 27

Lol okay okay....

Your first post on this...

"Select the appropriate restrictions for the variables."

And we found the restrictions to be 'a' can NOT equal -3

Is THIS the same type of problem? are we still "Select the appropriate restrictions for the variables." ?

Answer 28

Simplify and select the answer with the appropriate restrictions for the variables.


are the directions

Answer 29

Alright then yes we are doing the SAME thing as we did before ...okay?

We are finding what value of 'x' we can plug in to make our new denominator = 0

Our new denominator is 2x^4

So 2x^4 = 0 if we plug in x = 0....we get 0....so our restriction is x can NOT = 0

That is your answer :)

Answer 30

thats none of my answers though

Answer 31

Let me see your choices

Answer 32

Oh that's because you gave me a different question! :P

you gave me \(\huge \frac{5x - 3x^2}{2x^4}\)

But the question you are woking on is \(\huge \frac{(x+ 3)(x + 7)}{3x + 21} \)

Answer 33

ohh i did im sorry