Question

-3|2-5/4u|(less than or equal to sign) -18

Answer 1

yes .

Answer 2

Okay

Answer 3

You can continue solving for u from there

Answer 4

Is that what the back of your book says?

Answer 5

It says u <= -3 1/5 or u >= 6 2/5

Answer 6

Maybe you interpreted the problem wrong. can you scan it directly from the book? Isn't it in your Algebra book? I still have that. Tell me the section and page number.

Answer 7

Page 421, problem 17 .

Answer 8

Something is very weird about it because -3 1/5 = -16/5 and 6 2/5 = 32/5
It's like backwards or something

Answer 9

Maybe you can check it out .

Answer 10

I will

Answer 11

Chrissy, give me the exact page number because I'm having trouble finding it. Give me the actual book page number not the adobe page number, thanks

Answer 12

Remember, I'm using adobe, so..., but yeah, I can't find it, however, I'm in the right section, I just need you to double-check the page number

Answer 13

And also, tell me the name of the section of the book.

Answer 14

I will fix the solution

Answer 15

Okay .

Answer 16

\[-3|2-\frac{5u}{4}| \le -18\]

Answer 17

:(

I was going to fix the solution myininaya. It was an honest mistake where I put u in the denominator

Answer 18

\[|2-\frac{5u}{4}| \ge 6\]
\[|8-5u| \ge 24 \text{ note I multiplied both sides by 4}\]
\[|f(x)| \ge 24 => f(x) \ge 24 \text{ or } f(x) \le -24\]
So we have
\[8-5u \ge 24 \text{ or } 8-5u \le -24\]
\[-5u \ge 16 \text{ or } -5u \le -32\]
\[u \le \frac{-16}{5} \text{ or } u \ge \frac{32}{5}\]

Answer 19

I wish you had let me fix it

Answer 20

You can still do it. You want me delete this?

Answer 21

No, it's too late now.

Answer 22

I understand it . Thanks guys . :)