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arlik [135]
3 years ago
10

A rectangle and a square have the same area. a. The width of the rectangle is w inches. The length of the rectangle is 15 inches

less than 4 times the width of the rectangle. Write a polynomial to represent the area of the rectangle. b. The side of the square is w inches. Write a polynomial to represent the area of the square. c. Write an equation such that the area of the rectangle is equal to the area of the square. d. Solve the equation in part (c) to find the side length of the square. e. What are the length and the width of the rectangle?
Mathematics
1 answer:
harkovskaia [24]3 years ago
4 0

9514 1404 393

Answer:

  a) w(4w-15)

  b) w²

  c) w(4w -15) = w²

  d) w = 5

  e) 5 by 5

Step-by-step explanation:

a) If w is the width, and the length is 15 less than 4 times the width, then the length is 4w-15. The area is the product of length and width.

  A = w(4w -15)

__

b) If w is the side length, the area of the square is (also) the product of length and width:

  A = w²

__

c) Equating the expressions for area, we have ...

  w(4w -15) = w²

__

d) we can subtract the right side to get ...

  4w² -15w -w² = 0

  3w(w -5) = 0

This has solutions w=0 and w=5. Only the positive solution is sensible in this problem.

The side length of the square is 5 units.

__

e) The rectangle is 5 units wide, and 4(5)-15 = 5 units long.

The rectangle and square have the same width and the same area, so the rectangle must be a square.

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Please help!<br> And explain the answers too!
RideAnS [48]
1) He worked 4 hours leaving 6 more hours for him.
rate for A alone 6 baker
rate for A & B together 4
1/a+1/b=1/x
1/6+1/b=1/4
1/b=1/4-1/6
1/b=6/24-4/24
1/b=2/24
1/b=1/12
b=12
rate for wife alone 12 hours for the remaining 6 hours
check
1/6+1/12=1/4
12/72+6/72=1/4
18/72=1/4
ok
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But we want to know how long it would have taken her for the whole job not just part of it
he works twice as fast as she does so the job that takes him 10 hours alone would take her 20 hours alone.

2) jessica and cynthia work in a pet shop.
it takes jessica 6 hours to groom all the pets but cynthia needs 8 hours to groom them.
if jessica starts to groom 1 hour before cynthia joins, how long will it take them to finish?
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let t = time it takes them to finish (C's working time)
then
t+1 = J's working time
let 1 = completed job (all pets groomed)
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A shared work equation, each does a fraction of the job, the two fractions add up to 1
%28%28t%2B1%29%29%2F6 + t%2F8 = 1
Multiply by the least common multiple of 6 and 8, 24. cancel the denominators
4(t+1) + 3t = 24
4t + 4 + 3t = 24
4t + 3t = 24 - 4
7t = 20
t = 20/7
t = 26%2F7 hrs, which is: 2 + 6%2F7*60 = 2 hrs 51.43 min to finish the job

3) math club takes 40 minutes
science takes 50 minutes.
rate * time = quantity
quantity = 1 set up of the tables.
rate for math club is 1/40 of all the tables in 1 minute.
rate for science club is 1/50 of all the tables in 1 minute.
math club works for 10 minutes.
rate * time = quantity
1/40 * 10 = 10/40 = 1/4 of the tables are already set up by the math club.
there are 3/4 of the tables that still need to be set up.
rate * time = quantity
the rates of the math club and the science club are additive.
(1/40 + 1/50) * time = 3/4 of the tables that still need to be set up.
common denominator is 200.
(5/200 + 4/200) * time = 3/4.
9/200 * time = 3/4
time = 3/4 * 200/9 = 600 / 36 = 16 and 2/3 minutes.
total time required is 10 + 16 + 2/3 minutes = 26 + 2/3 minutes.
how to check.
10 minutes is used to do 1/4 of the tables.
16 + 2/3 minutes is used to do the remaining 3/4 of the tables.
combined rate of math and science club is 9/200 of the tables in 1 minute.
math club rate of 1/40 * 10 = 10/40 = 1/4 of the tables.
combine rate of 9/200 * (16 + 2/3) = 9/200 * 50/3 = 450/600 = 9/12 = 3/4 of the tables.
1/4 + 3/4 = 1 = all of the tables.

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