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Bumek [7]
3 years ago
14

From 5miles to 16miles

Mathematics
1 answer:
vfiekz [6]3 years ago
3 0
From 5 miles to 16 miles in 11 miles in between.
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The temperature in your town is 31°F. The radio announcer says that the temperature will drop 15 degrees. What will the temperat
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Answer:  16° F

Step-by-step explanation:

30-15=16

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5. Class 7887 has 2:7 math classes daily. How many math classes do they have
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56.7

Step-by-step explanation:

2.7 x 21 = 56.7

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(-9,5); m=4 right and equation in point slope form of the line that passes through the given point and with the given slope M
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Slope form of line  y = 4x +41

Step-by-step explanation:

m= 4

The y-intercept of the line that passes through the given point  (-9,5) is

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b= 5 +36

b=41

The equation of the line that passes through the given point  (-9,5) is

y =mx +b

 y = 4x +41

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Each cube in this rectangular prism is 1 cm3. What is the volume of the rectangular prism? A. 9 cm3 B. 20 cm3 C. 40 cm3 D. 48 cm
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A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a
Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 =  0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = 2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

W_2 =  \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5

The work done in pulling half the rope, W₂ = 562.5 lb·ft

6 0
3 years ago
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