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3 years ago

Write the expression as a polynomial (x+1)(x+2)(y–3)

1 answer:

6 0

$\text{Use FOIL method:}\ (a+b)(c+d)=ac+ad+bc+bd\\\\(x+1)(x+2)(y-3)=[(x)(x)+(x)(2)+(1)(x)+(1)(2)](y-3)\\\\=(x^2+2x+x+2)(y-3)=(x^2+3x+2)(y-3)\\\\=(x^2)(y)+(x^2)(-3)+(3x)(y)+(3x)(-3)+(2)(y)+(2)(-3)\\\\=\boxed{x^2y-3x^2+3xy-9x+2y-6}$

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2 years ago

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Step-by-step explanation:

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3 years ago

Get the general term for the sequence being your t3 = 11 and the t20 = 244.2

marusya05 [52]

Answer:

nth term = $t_{n} = 7.639(1.2)^{n - 1}$

Step-by-step explanation:

Let us assume that the given sequence is a G.P.

Now, if the first term of the G.P. is a and the common ratio is r, then

Third term = $t_{3} = ar^{2} = 11$ .......... (1) and

20th term = $t_{20} = ar^{19} = 244.2$ ........... (2)

Now, dividing equation (2) with equation (1) we get

$\frac{ar^{19} }{ar^{2} } = \frac{244.2}{11} = 22.2$

⇒ $r^{17} = 22.2$

⇒ r = 1.2.

Hence, from equation (1) we get

a(1.2)² = 11

⇒ a = 7.639 (Approx.)

Therefore, the general term of the sequence i.e. nth term = $t_{n} = 7.639(1.2)^{n - 1}$ (Answer)

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