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klemol [59]
2 years ago
7

The manager of a snack bar buys bottled water in packs of 35 and candy bars in packs of 20. Then, she sells the items individual

ly. Yesterday, she bought the same number of water bottles as candy bars to sell at the snack bar. If she bought the lowest number of items possible, how many packs of each item did she buy?
Mathematics
2 answers:
cestrela7 [59]2 years ago
8 0
4 packs of bottled water (35*4=140)
7 packs of candy bars (20*7=140)
Arte-miy333 [17]2 years ago
6 0

Answer:

4 packs of water and 7 packs of candy bars.

Step-by-step explanation:

We want to know the smallest number that both 35 and 20 will go into.  This is the least common multiple (LCM).

To find the LCM, we find the prime factorization of each number:

35 = 5(7)

20 = 5(4)

4 = 2(2)

20 = 5(2)(2)

We now multiply the common factor, 5, by all of the uncommon factors:

5(7)(2)(2) = 35(2)(2) = 70(2) = 140

This means she needs 140 bottles of water and 140 candy bars.

Water is sold in packs of 35; this means she needs

140/35 = 4 packs of water.

Candy bars are sold in packs of 20; this means she needs

140/20 = 7 packs of candy bars.

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Jamie's dog eats 3/4 pound of dog food each day. How many pounds of dog
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2. (15 points) Find the volume of the solid generated by revolving the region bounded by the curves x=
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Step-by-step explanation:

First, graph the region.  The first equation is x = 3y² − 2, which has a vertex at (-2,0).  The second equation is x = y², which has a vertex at (0, 0).  The two curves meet at the point (1, 1).  The region should look kind of like a shark fin.

(a) Rotate the region about y = -1.  Make vertical cuts and divide the volume into a stack of hollow disks (washers).

Between x=-2 and x=0, the outside radius of each washer is y₁ + 1, and the inside radius is 1.  Between x=0 and x=1, the outside radius of each washer is y₁ + 1, and the inside radius is y₂ + 1.

The thickness of each washer is dx.

Solve for y in each equation:

y₁ = √(⅓(x + 2))

y₂ = √x

The volume is therefore:

∫₋₂⁰ {π[√(⅓(x+2)) + 1]² − π 1²} dx + ∫₀¹ {π[√(⅓(x+2)) + 1]² − π[√x + 1]²} dx

∫₋₂⁰ π[⅓(x+2) + 2√(⅓(x+2))] dx + ∫₀¹ π[⅓(x+2) + 2√(⅓(x+2)) − x − 2√x] dx

∫₋₂¹ π[⅓(x+2) + 2√(⅓(x+2))] dx − ∫₀¹ π(x + 2√x) dx

π[⅙(x+2)² + 4 (⅓(x+2))^(3/2)] |₋₂¹ − π[½x² + 4/3 x^(3/2)] |₀¹

π(3/2 + 4) − π(½ + 4/3)

11π/3

(b) This time, instead of slicing vertically, we'll divide the volume into concentric shells.  The radius of each shell y + 1.  The width of each shell is x₂ − x₁.

The thickness of each shell is dy.

The volume is therefore:

∫₀¹ 2π (y + 1) (x₂ − x₁) dy

∫₀¹ 2π (y + 1) (y² − (3y² − 2)) dy

∫₀¹ 2π (y + 1) (2 − 2y²) dy

4π ∫₀¹ (y + 1) (1 − y²) dy

4π ∫₀¹ (y − y³ + 1 − y²) dy

4π (½y² − ¼y⁴ + y − ⅓y³) |₀¹

4π (½ − ¼ + 1 − ⅓)

11π/3

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The thickness of each washer is dy.

The volume is therefore:

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∫₀¹ π [(4 + y²)² − (3y² + 2)²] dy

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∫₀¹ π (-8y⁴ − 4y² + 12) dy

-4π ∫₀¹ (2y⁴ + y² − 3) dy

-4π (⅖y⁵ + ⅓y³ − 3y) |₀¹

-4π (⅖ + ⅓ − 3)

136π/15

5 0
3 years ago
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