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Reil [10]
3 years ago
11

The diagram shows a 3 cm x 5 cm x 4 cm cuboid.

Mathematics
1 answer:
kogti [31]3 years ago
3 0

Answer:

a) 5.83 cm

b) 34.4 deg

Step-by-step explanation:

a)

AC is the hypotenuse of a right triangle with legs measuring 3 cm and 5 cm.

c^2 = a^2 + b^2

c^2 = 3^2 + 5^2

c^2 = 9 + 25

c^2 = 34

c = sqrt(34) cm = 5.83 cm

b)

Triangle ACD is a right triangle with right angle DAC.

AD = 4 cm

AC = 5.83 cm

tan <ACD = opp/adj

tan <ACD = AD/AC

tan <ACD = 4/5.83

m<ACD = tan^-1 (0.68599)

m<ACD = 34.4 deg

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This implies that:

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So, we have:

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Integrate

V = 2\pi  * [{\frac{8x^2}{2} - \frac{x^{1+\frac{5}{2}}}{1+\frac{5}{2}}]\vert^4_0

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V = 2\pi  * [{4x^2 - \frac{x^{\frac{7}{2}}}{\frac{7}{2}}]\vert^4_0

V = 2\pi  * [{4x^2 - \frac{2}{7}x^{\frac{7}{2}}]\vert^4_0

Substitute 4 and 0 for x

V = 2\pi  * ([{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}] - [{4*0^2 - \frac{2}{7}*0^{\frac{7}{2}}])

V = 2\pi  * ([{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}] - [0])

V = 2\pi  * [{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}]

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V = 2\pi  * [{64 - \frac{2}{7}*2^7]

V = 2\pi  * [{64 - \frac{2}{7}*128]

V = 2\pi  * [{64 - \frac{2*128}{7}]

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V = 2\pi  * [\frac{64*7-256}{7}]

V = 2\pi  * [\frac{448-256}{7}]

V = 2\pi  * [\frac{192}{7}]

V = [\frac{2\pi  * 192}{7}]

V = \frac{\pi  * 384}{7}

V = \frac{384}{7}\pi

Hence, the required volume is:

Volume = \frac{384}{7}\pi

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The possible products are:
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