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Solnce55 [7]
3 years ago
9

Wholesale Distributors sells to its contractors with a 42% markup on cost. If the selling price for cabinets is $9,182, what is

the cost to contractors based on cost? (Round your answer to the nearest cent.)
Mathematics
1 answer:
trasher [3.6K]3 years ago
6 0
Selling price = cost + 0.42*cost = 1.42*cost
cost = selling price/1.42

cost = $9182/1.42 ≈ $6466.20
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The regular price of an item is $84. The item is on sale witha discount of $12. What is the discount rate.
sammy [17]

Answer:

Discount = 14.3% (rounded up to one decimal place)

Step-by-step explanation:

The regular price of an item is $84

Discount = $12

Percentage discount = (Discount ÷ regular price) × 100

Percentage discount = \frac{12}{84} × 100 = 14.3% (rounded up to one decimal place)

5 0
3 years ago
What are the intercepts???
IRINA_888 [86]

Answer:

X is intercepted at -7

Y is intercepted at 2

Step-by-step explanation:

5 0
3 years ago
The length of a rectangle is 5 more than the side of a square, and the width of the rectangle is 5 less than the side of the squ
ANEK [815]

Answer:

20

Step-by-step explanation:

Given :

Length = l

Width = w

Area of square = 45

Let side of square = x

Area of square = x²

45 = x²

√45 = x

l of rectangle = x + 5 = √45 + 5

w of rectangle = x -5 = √45 -5

Area of rectangle= l * w

Area = (√45 + 5) * (√45 - 5)

Area = 45 - 5√45 + 5√45 -25

Area = 45 -25

Area of rectangle = 20

6 0
3 years ago
HELP ASAP
Liula [17]

Answer:

A = 1.5386

Step-by-step explanation:

Quarter circle area formula:

A = (πr²)/4

A= (π1.4²)/4

A = (3.14*1.96)/4

A = (6.1544)/4

A = 1.5386

7 0
2 years ago
Read 2 more answers
A box with a square base and an open top is being constructed out of A cm2 of material. If the volume of the box is to be maximi
viktelen [127]

Answer:

Side length = \sqrt{\frac{A}{3} } cm ,   Height =  \frac{1}{2} \sqrt{\frac{A}{3} } cm  ,  Volume = \frac{A\sqrt{A}}{6\sqrt{3} }  cm³

Step-by-step explanation:

Assume

Side length of base = x

Height of box = y

total material required to construct box = A ( given in question)

So it can be written as

A = x² + 4xy

4xy = A - x²

  1. y = \frac{A - x^{2} }{4x}

Volume of box = Area x height

V = x² ₓ y

V = x² ₓ ( \frac{A - x^{2} }{4x} )

V =  \frac{Ax - x^{3} }{4}

To find max volume put V' = 0

So taking derivative equation becomes

\frac{A - 3 x^{2} }{4} = 0

A = 3 x^{2}

x^{2} = \frac{A}{3}

x = \sqrt{\frac{A}{3\\} }

put value of x in equation 1

y = \frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }  

y = \frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }

y = \frac{1}{2} \sqrt{\frac{A}{3} }

So the volume will be

V = x^{2} × y

Put values of x and y from equation 2 & 3

V = \frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )

V = \frac{A\sqrt{A}}{6\sqrt{3} }

8 0
4 years ago
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