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zzz [600]
3 years ago
6

If f(x) = –4x2 – 6x – 1 and g(x) = –x2 – 5x + 3, find (f – g)(x).

Mathematics
2 answers:
nexus9112 [7]3 years ago
8 0

Answer:

(f – g)(x) = -3x² - x -4.

Step-by-step explanation:

Given : If f(x) = –4x² – 6x – 1 and g(x) = –x²– 5x + 3.

To find :  (f – g)(x).

Solution ; We have given that  f(x) = –4x² – 6x – 1

g(x) = –x²– 5x + 3.

(f – g)(x) = f(x) - g(x).

(f – g)(x) = (–4x² – 6x – 1 ) - ( –x²– 5x + 3) .

(f – g)(x) = –4x² – 6x – 1 + x²+ 5x - 3 .

Combine like terms

(f – g)(x) = –4x² +  x² - 6x + 5x -1 -3.

(f – g)(x) = -3x² - x -4.

Therefore,  (f – g)(x) = -3x² - x -4.

Leokris [45]3 years ago
5 0
(-4x^2 -6x -1) - (-x^2 -5x +3) = -3x^2 - x - 4
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Read 2 more answers
QT 1.
erma4kov [3.2K]
1: 3.2
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A coin is tossed 6 times. what is the probability that the number of heads obtained will be 2? express your answer as a fraction
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3 years ago
Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1
defon
I'm guessing the series is supposed to be

\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}

By the ratio test, the series converges if the following limit is less than 1.

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|

The first n terms in the numerator's denominator cancel with the denominator's denominator:

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|

|x^n| also cancels out and the remaining factor of |x| can be pulled out of the limit (as it doesn't depend on n).

\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0

which means the series converges everywhere (independently of x), and so the radius of convergence is infinite.
3 0
3 years ago
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