For the homework, on pg 133 question 2B how is the answer x=6? I got all real numbers but the algebra solutions say that it is only 6.
Because it says and not or.
how do you solve 2c on page 145?_ _ \_(>_<)_/ I am so confused...??? | _/ \_
Will you actually look at my answer this time?
yes ma'am
Multiply both side by the denominator and get all the variables on one side. What do you get?
Similar to #11
Did you get it?
x=(2x)(7-x)
well i got both answers but using two different equations bc i did it twice
Multiply it out and move the variables to one side
0=14x-2xsquared-x
13x-2xsquared=0
Wouldn't it be 0 = 13x - x^2
Using reverse distribution. X(13-x) =0
why?
It would help if you hit reply to which one you are saying why to or being more specific.
I'm still confused...Can I come in tomorrow at lunch?
Advisory. Lunch I have a meeting
Post a Comment
19 comments:
For the homework, on pg 133 question 2B how is the answer x=6? I got all real numbers but the algebra solutions say that it is only 6.
Because it says and not or.
how do you solve 2c on page 145?
_ _
\_(>_<)_/ I am so confused...???
|
_/ \_
Will you actually look at my answer this time?
yes ma'am
Multiply both side by the denominator and get all the variables on one side. What do you get?
Similar to #11
Did you get it?
x=(2x)(7-x)
well i got both answers but using two different equations bc i did it twice
Multiply it out and move the variables to one side
0=14x-2xsquared-x
13x-2xsquared=0
Wouldn't it be 0 = 13x - x^2
Using reverse distribution. X(13-x) =0
why?
It would help if you hit reply to which one you are saying why to or being more specific.
I'm still confused...Can I come in tomorrow at lunch?
Advisory. Lunch I have a meeting
Post a Comment