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Nonamiya [84]
3 years ago
7

PLEASE HELP ME!!!!!!!!!

Mathematics
1 answer:
mihalych1998 [28]3 years ago
3 0

Answer:

A) 24

Step-by-step explanation:

Total weight - loaded weight = space remaining in the trailer: 1050kg - 82kg = 968kg

space remaining ÷ the number of transport: 968 ÷ 40 = 24.2 OR 24

Hope it helps you

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Eva is picking fruit at a farm. The fees are $5 for entry to the farm, $2.50 per pound of blueberries, and $2 per pound of apple
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Answer:

The required expression is: y = 5 + 2.50b + 2a

Step-by-step explanation:

Given

Entry = \$5

Blueberry = \$2.50

Apple = \$2

Required

Write an expression for this scenario

If 1 pound of blueberry cost $2.50

then b pounds = 2.50 * b

If 1 pound of apple cost $2

then a pounds = 2 * a

The required expression (y) is then calculated as

y = Entry\ Fee + Blueberry = Apple

y = 5 + 2.50 * b + 2 * a

y = 5 + 2.50b + 2a

Hence, the required expression is: y = 5 + 2.50b + 2a

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Question 6 of 10 Multiple Choice: Please select the best answer and click "submit." What is the greatest number of acute angles
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Jorge is asked to build a box in the shape of a rectangular prism. The maximum girth of the box is 20 cm. What is the width of t
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Answer:

The width of the box is 6.7 cm

The maximum volume is 148.1 cm³

Step-by-step explanation:

The given parameters of the box Jorge is asked to build are;

The maximum girth of the box = 20 cm

The nature of the sides of the box = 2 square sides and 4 rectangular sides

The side length of square side of the box = w

The length of the rectangular side of the box = l

Therefore, we have;

The girth = 2·w + 2·l = 20 cm

∴ w + l = 20/2 = 10

w + l = 10

l = 10 - w

The volume of the box, V = Area of square side × Length of rectangular side

∴ V = w × w × l = w × w × (10 - w)

V = 10·w² - w³

At the maximum volume, we have;

dV/dw = d(10·w² - w³)/dw = 0

∴ d(10·w² - w³)/dw = 2×10·w - 3·w² = 0

2×10·w - 3·w² = 20·w - 3·w² = 0

20·w - 3·w² = 0 at the maximum volume

w·(20 - 3·w) = 0

∴ w = 0 or w = 20/3 = 6.\overline 6

Given that 6.\overline 6 > 0, we have;

At the maximum volume, the width of the block, w = 6.\overline 6 cm ≈ 6.7 cm

The maximum volume, V_{max}, is therefore given when w = 6.\overline 6 cm = 20/3 cm  as follows;

V = 10·w² - w³

V_{max} = 10·(20/3)² - (20/3)³ = 4000/27 = 148.\overline {148}

The maximum volume, V_{max} = 148.\overline {148} cm³ ≈ 148.1 cm³

Using a graphing calculator, also, we have by finding the extremum of the function V = 10·w² - w³, the coordinate of the maximum point is (20/3, 4000/27)

The width of the box is;

6.7 cm

The maximum volume is;

148.1 cm³

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3 years ago
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