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myrzilka [38]
2 years ago
12

Help me please I’m giving briainly need help ASAP

Mathematics
1 answer:
Helen [10]2 years ago
6 0

Answer:

f(-x) = -4x^3 + 9x, f(x) is an odd function

Step-by-step explanation:

We are given that f(x) = 4x^3 - 9x. In order to solve this problem, we first need to understand what an odd function is and what an even function is. When we have an odd function, f(-x) = -f(x). Meaning, that if we plug -x in for x, we should get -f(x). On the other hand, with an even function, f(-x) = f(x). Even if you change the sign, when you simplify, you will get back the same function. In either case, we have to first find f(-x) to figure this problem out:

f(-x)=4(-x)^3-9(-x)\\f(-x)=-4x^3+9x

We can see from our answer for f(-x) that it is equal to -f(x). We can verify this below:

-f(x)=-(4x^3-9x)\\-f(x)=-4x^3+9x

Since f(-x) = -f(x), our function f(x) is an odd function

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Answer:

<em>Correct choices: a) 5:12 and c) 7 to 12</em>

Step-by-step explanation:

The ratio of red beads to pink beads is 5:7. This means that for every 5 red beads there are 7 pink beads.

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Answer:

Step-by-step explanation:

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3 years ago
If a sequence c1,c2,c3,...has limit K then the sequence ec1,ec2,ec3,...has limit e^K. Use this fact together with l'Hopital's ru
aleksandrvk [35]

Answer:

Step-by-step explanation:

If a sequence c1,c2,c3,...has limit K then the sequence ec1,ec2,ec3,...has limit e^K. Use this fact together with l'Hopital's rule to compute the limit of the sequence given by

bn=(n)^(5.6/n).

a)

L =  \lim_{n \to \infty} b_n \\\\\\L= \lim_{n \to \infty} n^{\frac{5.6}{n} }

Log on both sides

In (L) =  \lim_{n \to \infty} In (n)^{\frac{5.6}{n} }\\\\= \lim_{n \to \infty} \frac{5.6}{n} In(n)

=5.6 \lim_{n \to \infty} \frac{d}{dn}  In(n)/\frac{d}{dn} (n)\\\\=5.6 \lim_{n \to \infty} \frac{1}{n} /1 \\\\=5.6 \lim_{n \to \infty} \frac{1}{n} \\\\=5.6 \times 0\\\\In(L) =0\\\\L=e^0\\\\L=1

\therefore  \lim_{n \to \infty} (n)^{\frac{5.6}{n} =1

5 0
3 years ago
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