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TEA [102]
1 year ago
8

Ten turns of a wire are helically wrapped around a cylindrical tube with outside circumference 4 inches and length 9 inches. The

ends of the wire coincide with ends of the same cylindrical element. Find the length of the wire. (a) 41 inches (b)36 inches (c) 40 inches (d) 90 inches
Mathematics
1 answer:
kari74 [83]1 year ago
8 0

Answer:

b) 36 inches

Step-by-step explanation:

Length  of the  wire = Outside circumference of the cylindrical tube * length of the cylinder

= 4 * 9

= 36 inches

Length of the wire will be same to the surface area of the cylinder

Surface area of cylinder = circumference * length

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Answer:

(D)\left(-\dfrac38\right)\left(-\dfrac57\right)\left(-\dfrac14\right)

Step-by-step explanation:

The given options are:

(A)\left(-\dfrac38\right)\left(-\dfrac57\right)\left(\dfrac14\right)\\(B)\left(\dfrac38\right)\left(-\dfrac57\right)\left(-\dfrac14\right)\\(C)\left(\dfrac38\right)\left(\dfrac57\right)\left(\dfrac14\right)\\(D)\left(-\dfrac38\right)\left(-\dfrac57\right)\left(-\dfrac14\right)

The key to determining which product is negative is to understand the rule of sign multiplication.

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In Options A and B, the number of negative signs is even, therefore our result is positive.

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In Option D, the number of negative signs is odd, therefore our result is negative.

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1 year ago
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vovangra [49]

Answer: Option A) 729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12} is the correct expansion.

Explanation:

on applying binomial theorem,  (a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r

Here a=3c, b=d^2 and n=6,

Thus, (3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r

⇒ (3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6

⇒(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}

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