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Sophie [7]
2 years ago
9

Sections of prefabricated fencing are each 4 1/3 feet long. how long are 6 1/2 sections placed end to end

Mathematics
2 answers:
zalisa [80]2 years ago
6 0

we know that

4\frac{1}{3}= \frac{(4*3+1)}{3} =\frac{13}{3}

6\frac{1}{2}= \frac{(6*2+1)}{2}=\frac{13}{2}

To find how long are 6\frac{1}{2}  sections placed end to end

Multiply the number of sections by the length of one section

so

\frac{13}{2} *\frac{13}{3} =\frac{169}{6}\ ft

\frac{169}{6}\ ft =28\frac{1}{6}\ ft

therefore

<u>the answer is</u>

28\frac{1}{6} \ ft


ladessa [460]2 years ago
6 0
Each section is 4 1/3 feet long.
6 1/2 section will be 4 1/3 x 6 1/2 = 13/3 x 13/2 = 169/6 = 28 1/6 feet

Thus, 6 1/2 sections will be 28 1/6 feet long.
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dezoksy [38]

Answer:

a). Cost of 44 pupils = $14265

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Step-by-step explanation:

The given question is incomplete; here is the complete question.

The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.

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Let the equation representing the total cost of maintaining a school is,

C = ax + b

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    Equation will be,

    15705 = 50a + b -------(1)

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    13305 = 40a + b --------(2)

    By subtracting equation (2) from equation (1),

    15705 - 13305 = (50a + b) - (40a + b)

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    b = 3705

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    C = 240x + 3705

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    Total fee of 'p' pupils = $360p

    Total cost to run the school will be = 3705 + 240p

    For the school not to be in the loss,

    360p ≥ 3705 + 240p

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    p ≥ \frac{3705}{120}

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2 years ago
Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 1 + sec
masha68 [24]

Using the washer method, the volume is given by the integral

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where 3 - 1 = 2 is the distance from <em>y</em> = 3 to the axis of revolution, and similarly (1 + sec(<em>x</em>)) - 1 = sec(<em>x</em>) is the distance from <em>y</em> = 1 + sec(<em>x</em>) to the axis. The integrand is symmetric about <em>x</em> = 0, so the integral "folds" in on itself, and the integral from -π/3 to π/3 is twice the integral from 0 to π/3.

So the volume is

\displaystyle2\pi\int_0^{\pi/3}(4-\sec^2x)\,\mathrm dx=2\pi(4x-\tan x)\bigg|_0^{\pi/3}=\boxed{\dfrac{8\pi^2}3-2\pi\sqrt3}

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2 years ago
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