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blondinia [14]
2 years ago
10

Thomas is saving money for a new mountain bike. The amount (a) Thomas needs to save is more than $50.45. Which inequality models

the amount Thomas needs to save?
A.$50.45=a
B.$60.45 C.a>$50.45
D.a<$50.45
Mathematics
1 answer:
AfilCa [17]2 years ago
3 0
C.

This is because the amount is more than what he needs to save, considering that he is probably has some money in his bank already.
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Which of the following is the expansion of (3c + d2)6?
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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}

⇒(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}

⇒(3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}

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V=\left(\frac{d-2}{3d-9}\right)\left(\frac{4}{d-4}\right)\left(\frac{2d-6}{2d-4}\right) \\ \\ \\ Simplifying: \\ \\ V=\frac{d-2}{3d-9}\cdot \frac{4}{d-4}\cdot \frac{d-3}{d-2} \\ \\ V=\frac{\left(d-2\right)\cdot \:4\left(d-3\right)}{\left(3d-9\right)\left(d-4\right)\left(d-2\right)} \\ \\ V=\frac{4\left(d-3\right)}{\left(3d-9\right)\left(d-4\right)} \\ \\ V=\frac{4\left(d-3\right)}{3\left(d-3\right)\left(d-4\right)}

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