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Alla [95]
1 year ago
8

The maximum load of a horizontal beam that is supported at both ends varies jointly as the width and the height and inversely as

the length between the supports. A beam 6m long, 0.1m wide, and 0.06 m high supports a load of 360kg. What is the maximum load supported by a beam 16m long, 0.2m wide, and 0.08 m high?
Mathematics
1 answer:
dezoksy [38]1 year ago
8 0

Answer:

360kg

Explanation:

Here, we want to get the maximum load supported

From the first part of the question, we have it that:

The maximum load g, varies jointly with the width w and the height h, and inversely as the length l between the supports

Mathematically, we have that as:

g\text{ = k }\times\frac{wh}{l}

where k is the proportionality constant

Now, let us get the value of k

We substitute the first set of values as follows:

\begin{gathered} 360\text{ = }\frac{k\times0.1\times0.06}{6} \\  \\ k\text{ = 360,000 }\frac{kg}{m} \end{gathered}

Now, using this value of k, we want to get the value of g

Mathematically, we have that as:

g\text{ = }\frac{360,000\times0.2\times0.08}{16}\text{ = 360 }kg

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Answer:

9.5%

Step-by-step explanation:

Buying price of the part= $ 8.75

Selling price of the part= $ 9.98

sale tax= $0.40

Selling price before sale tax= $9.98-$ 0.40 = $ 9.58

Make-up amount charged= $9.58- $8.75 = $ 0.83

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=9.5%

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Describe a rectangular prism using words such as surfaces
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Which hundredth is 8.275 beetween
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Answer:

the 7 is in the hundredths place

Step-by-step explanation:

hope this helps :)

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3 years ago
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A square has a perimeter of 28 feet what is its area
Westkost [7]
Given:
square shape
Perimeter 28 feet
Find its area

A square has 4 equal sides, so perimeter is 4a

Perimeter = 4a
28 ft = 4 a
28 ft / 4 = a
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One side of the the square is 7 ft.

Area of a square is the measure of one side raised to the power of 2.

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7 0
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A bank sampled its customers to determine the proportion of customers who use their debit card at least once each month. A sampl
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Answer:

(0.084,0.396)

Step-by-step explanation:

The 99% confidence interval for the proportion of customers who use debit card monthly can be constructed as

p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }

p=\frac{x}{n}

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0.084

The 99% confidence interval for the proportion of customers who use debit card monthly is (0.084,0.396).

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