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Nat2105 [25]
4 years ago
11

A candle maker has 4 1/2 pounds of clear wax. He wants to cut the wax into pieces that are 2/3 pound each. How many 2/3 pound pi

eces can he divide the wax into? How much wax is left over?
Mathematics
1 answer:
prisoha [69]4 years ago
8 0

Answer:

a)  10  pieces of  2/3 pound  can be made.

b) 0.54 pounds of wax is left over.

Step-by-step explanation:

Total amount of clear wax available =  4\frac{1}{2}  = \frac{9}{2} pounds

So, the maker has in total 4.5 pounds of wax

Now, The weight of each piece = \frac{2}{3}  = 0.66  pounds

To find the number of pieces that can be made out of 4.5 pounds of wax,

let that number = n

So, n = \frac{\textrm{Total weight of the clear wax}}{\textrm{Weight of each wax}}

or, n = n = \frac{4.5}{0.66}  = 6.81

So, nearly 6 pieces can be made completely out of the clear wax

Left over amount of wax  = Total weight -  Weights of the pieces made

                                            =  4.5- 6(0.66)

                                            = 0.54 pounds

Hence, 0.54 pounds of wax is left over

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An indoor sport exhibition is coming to the arena. Your supervisor has asked you to help set up a handball pitch and seating are
Sphinxa [80]

Answer:

Perimeter: 174.8 m

Area: 1,394 sq m

Step-by-step explanation:

First, the perimeter.

Before we start, let's calculate the circumference of the half-circles at the ends of the field.

The measurement says 2,000 cm, so let's convert it to 20 m for ease.

Circumference of a circle: πd, where d = diameter.. in our case d = 20 m

Circumference of a 20m diameter circle: 20π = 62.8 m

We have 2 half circles... so the perimeter of each half-circle will be: 31.4 m

We also have 800 cm measurement for the "height" of the seating areas... let's convert that in 8 m

We also need to find out the space between the seating area...  We know the whole rectangular pitch is 40 m, then we have to subtract the width of both seating areas (20 and 15 m)... so the space between them is 5m

So, starting with the upper left corner of the rectangular pitch, and working our way clockwise, we encounter the following lengths:

P = 40 + 31.4 + 8 + 10 + 8 + 5 + 8 + 25 + 8 + 31.4 = 174.8 m

The total perimeter is then of 174.8 m

For the area, we need to calculate the area of all forms:

Large rectangular pitch:

LR = 40 x 20 = 800 sq m

The two half circles, form a circle, so A = πr², where r is the radius, which is half the diameter.

AC = π (10)² = 100 π = 314 sq m

Then the seating areas:

SA1 = 25 x 8 = 200 sq m

SA2 = 10 x 8 - 80 sq m

Then, we add up everything:

TA = LR + AC + SA1 + SA2

TA = 800 + 314 + 200 + 80 = 1,394 sq m

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