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1 year ago

Evaluate the function g(x) = –2x2 + 3x – 5 for the input values –2, 0, and 3

Mathematics

2 answers:

Verizon [17]1 year ago

8 0

Answer: g(-2) = -15, g(0) = -9,

g(3) = -18

Step-by-step explanation:

g(-2) = -2*2 + 3(-2) - 5

g(-2) = -4 - 6 - 5

g(-2) = -15

g(0) = -4 - 0 - 5

g(0) = -9

g(3) = -4 - 3(3) - 5

g(3) = -4 - 9 - 5

g(3) = -18

Sunny_sXe [5.5K]1 year ago

4 0

Answer:

g(x)= -19

g(x)= -5

g(x)= -14

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2 years ago

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.

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

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

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

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