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.

When 4 inches are added to each side of square, its area increases by 64 square inches. What is the area of the initial square?

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

36
\[A=x ^{2}\] with x being the length of the initial sides \[A+64=(x+4)\]
the area of previous square was 36 sq. inches

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[(x+4)^{2}\] sorry
(x + 4)^2 = x^2 + 64. This will be a quadratic in "x".
how did you get 36 so i can do the rest by myself?
it can be done by trial and error also
(x + 4)^2 = x^2 + 64 will become x^2 + 8x + 16 = x^2 + 64. Just combine terms and simplify.
ok one thing why do you square it
area of square is its side squared
You have to get the area. Both of the first square and the second.
and x is the side
thanks guys
The second area is (x + 4)^2 and the first area is x^2
hmm
its our pleasure
\[A _{1}=x _{1}^{2}\] \[A _{2}=x _{2}^{2}\] \[A _{2}=A _{1}+64\] \[x _{2}=x _{1}+4\] Does that help?
lol i got it
ok so i did x^2 + 8x + 16 = x^2 + 64. and i got 6 what did i do wrong
you got \[x _{1}=6\] so what's the area? Never forget what was the initial question

Not the answer you are looking for?

Search for more explanations.

Ask your own question