Hi! Could you type the question out.
Sqrt(1-3x)=x+3
[sqrt(1-3x)]^2=(x+3)^2
1-3x=(x+3)(x+3)
1-3x=x^2+6x+9
-3x=x^2+6x+8
0=x^2+9x+8
The answers are -1 and -8 BUT, we have to plug them back into the original equation to make sure we don't get a negative under the square root sign.
After doing this, we realize that only -1 works, so the answer is x=-1
Sorry that this took forever to answer. I was thinking of a good way to explain this, and if you need any further explanation, message me:)
Best wishes:)
All we have to do is set this up!
we know that all he practiced altogether for 15 and 1/3 hours
so lets put that down so far
1
15 ----- =
3
and he says that the first week he played 6 and 1/4 hours on the first week
now lets add that
1 1
15 ----- = 6 ----- +
<span> 3 4
</span>now we also know that he played 4 and 2/3 hours on the first week
now lets add that
1 1 2
15 ----- = 6 ----- + 4 ----- +
<span> 3 4 3
</span>we arent finished yet!
now the third week we dont know how long he played so we are going to put x in its place
1 1 2
15 ----- = 6 ----- + 4 ----- + x
<span> 3 4 3
</span>and there is your answer! finally...
hope this helps:) MARK AS BRAINLIEST!!!
:D
Answer:
A. 0
E. -3
F. 9
Step-by-step explanation:
You can't divide by 0; it is undefined. So if x cannot equal zero, then anything that turns the denominator to zero is an asymptote. Therefore, the roots of the cubic expression would be excluded, and we get our final answers.
