Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Find a and b. If |x-3|<7, then a

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

-7
If you were to substitute in -4 and10 in to a
Are we doing piece wise functions here?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I do not understand? Piece wise functions?
You haven't learned that? Then I got my answer.
LOL! Probably not, still in the beginning of College Algebra
This is not college algebra.
This is Algebra 2.
That makes senses, but it is called College Algebra at my school, and towards end of semester it gets in to Trig
That is algebra two. And this is piece wise functions. I am going to go offline for a while be back in about ten or fifteen minutes.
Thank you
\(\bf |x-3|<7 \implies \begin{cases} +(x-3)<7\\ \quad \\ \bf -(x-3)<7 \end{cases}\) if you were to solve those 2 cases, or scenarios, what would be the "values" of "x"?
I am so lost
hmm ok, what part?
When I work those problems, which I am probably not doing correctly I get:\[+(x−3)<7 = x<10\] \[−(x−3)<7 = x>4\] but the answer is -4, so I believe I am doing this wrong
So would I have multiplied both sides by -1 and then added three to both sides?
those answers are correct
now to make each of the inequalities look like " x + 4 " just add 4 to each side :)
hmm... the 2nd one would be -4l
other than that, is ok... so \(\large {|x-3|<7 \implies \begin{cases} +(x-3)<7 \implies x < 10\\ \quad \\ -(x-3)<7 \implies x > -4 \end{cases}\\ \quad \\ x < 10 \implies x +4 < 14\\ \quad \\ x > -4 \implies x+4 > 0}\)
so you can see what "a" and "b" are
So I did end up with -4
\(\large {\begin{cases} x < 10 \implies x +4 < 14\\ \quad \\ x > -4 \implies x+4 > 0 \end{cases}\implies \bf 0 < x +4 < 14}\)
So I would add 4 to both sides and not subtract?
well, they want \(\large x \color{red}{+}4\) so you'd need to add, yes
-4
yeap when yo have only left and right sides, you add/divide/multiply to BOTH sides when you have left, middle, and right sides, you add/divide/multiply to ALL sides
-4
\[or we can say a \le 0,b \ge 14 .\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question