The width of a rectangle is 6 in. less than its length. The perimeter is 68 in. What is the width of the rectangle? in.

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 our expert's

answer on brainly

SEE EXPERT ANSWER

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

A community for students.

The width of a rectangle is 6 in. less than its length. The perimeter is 68 in. What is the width of the rectangle? in.

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

Hi pooja.
Uh i think the answer is in the question....if im right
divide it in two first

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1442028600183:dw|
Omg I forgot the shapes have the "fill" option.... anyway.
yeah
When you say divide two first, do you mean 68 ÷ 2 and 6÷ 2
no 68 by 2
then subtract 6 from that
okay, that is 34...
the length of the rectangle is given, but they say the `width` `is (equal to)` the `length - 6 in`You can rewrite this as \(\sf w = L-6\)
So the Perimeter of a rectangle is represented as: \[\sf P = 2L +2W\]
Since we're given the perimeter, and that is = 68 in. we can substitute that in to our formula. \[\sf 68 = 2L +2W\]
But we also know something else, \(\sf W=L-6\)
68 - 6?
So we can substitute this in to the formula for the perimeter. \[\sf 68 = 2L+2(L-6)\]
Are you following, @gabbyalicorn ?
Um, somewhat... I don't really get it but i'm following the steps if that's what you mean. :)
Which part are you confused with? Let's clarify that before moving on.
is L the variable we are trying to find
the number for...
We are trying to find W (width). The problem indiscreetly gives us the equation for the width by saying that the width IS 6 LESS THAN the length. We can write an equation for that. \(\sf w = L-6\)
but what is L
But we can't automatically find the width, \(\sf w\) can we? We need to find the length, \(\sf L\) first.
That is where we use the equation for the perimeter of the rectangle. \(\sf P = 2L +2w\)
Do you see how this works?
I think I'm getting it...
Okay, so we have \[\sf 68 = 2L+2(L-6)\] We need to first distribute (meaning multiply) 2 to each term inside the parenthesis. \[2 \cdot L =~?\]\[2\cdot (-6) =~?\]
|dw:1442029733818:dw|
Good
:)
\[\sf 68 = 2L +2L-12\]We add +12 to both sides of the equation.
why - 12 I thought you said add
because we want to isolate the L's to one side of the equation, we need to eliminate the 12 from that side. In order to eliminate it we need to add +12 to it. \[\sf +12-12 = 0\]So, \(\sf 68 + 12 =~?\)
80
So now we have \[\sf 80 = 2L + 2L\]\[\sf 2L + 2L =~?\]
4L ?
Good.
:}
\[\sf 80 = 4L\]Now we want to isolate the right side of the equation so we have JUST L. This is when we divide BOTH sides of the equation by 4. \[\sf \frac{80}{4}=~?\]
20
ALRIGHT! So now we have figured out \(\sf L\). \[\sf L=20\]Now we can go back to our equation for width and plug in L. \[\sfW=L-6\]So therefore what does W =?
Sorry, typo. \(\sf W=L-6\)*
20 - 6 = 14?
That's correct :)
Yes. :)
Great! You worked it out yourself :)
:D Thank you for helping me and not giving up on me! I reallu appreciate it. :)
No problem ^^

Not the answer you are looking for?

Search for more explanations.

Ask your own question