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tatiyna
4 years ago
14

10p+2-2p=4p=10 a) 1 b)-1 c) 3 d)-3

Mathematics
1 answer:
krek1111 [17]4 years ago
6 0
If 4p is equal to 10 then p should be equal to 10\4=2.5 but the option is not there
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Answer:

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Nicholas needs to write down the 6 and "carry" the 4, adding it to the numbers of tenths.

Depending on the method you've been taught, the 4 may be written above the column with place value of tenths, or it may be written on a separate sum line.

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A sheet of paper 90 cm-by-66 cm is made into an open box (i.e. there's no top), by cutting x-cm squares out of each corner and f
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26 - \sqrt{181} cm

Step-by-step explanation:

The volume of the box is:

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V = (66*x - 2*x^2)*(90 - 2*x)

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V = 4*x^3 - 312*x^2 + 5940*x

where x is the length of the sides of the squares,  in cm.

The mathematical problem is :

Maximize: V = 4*x^3 - 312*x^2 + 5940*x

subject to:

x > 0

2*x < 66 <=> x < 33

In the maximum, the first derivative of V, dV/dx, is equal to zero

dV/dx = 12*x^2 - 624*x + 5940

From quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)}

x = \frac{624 \pm \sqrt{(-624)^2 - 4(12)(5940)}}{2(12)}

x = \frac{624 \pm \sqrt{104256}}{24}

x = \frac{624 \pm \sqrt{2^6*3^2*181}}{24}

x = \frac{624 \pm 8*3*\sqrt{181}}{24}

x_1 = \frac{624 + 24*\sqrt{181}}{24}

x_1 = 26 + \sqrt{181}

x_2 = \frac{624 - 24*\sqrt{181}}{24}

x_2 = 26 - \sqrt{181}

But x_1 > 33, then is not the correct answer.

5 0
3 years ago
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