Number of real solutions? 8x^2 - 11x = -3 -4x^2 - 4 = 8x

Mathematics

1 answer:

Anna71 [15]3 years ago

3 0

(1) ANSWER

$x = \frac{3}{8} \: or \: x = 1$

(2) ANSWER

$x = - 1$

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Consider the following function. f(x) = 9 − x2/3 Find f(−27) and f(27). f(−27) = f(27) = Find all values c in (−27, 27) such tha

oksano4ka [1.4K]

I guess the function is $f(x)=9-x^{2/3}$ . Then $f(-27)=0$ and $f(27)=0$ .

The derivative is $f'(x)=-\dfrac23 x^{-1/3}$ , but there is no $c$ such that

$-\dfrac23c^{-1/3}=0$

This doesn't contradict Rolle's theorem because $f'(0)$ does not exists. In other words, $f$ is not differentiable on (-27, 27), so the conditions of Rolle's theorem are not met. (Looks like that would be the last option, or the second to last option if the last one is "Nothing can be concluded")

6 0

3 years ago

Will give, Branliest.

Vlad [161]

Answer:

Step-by-step explanation:

x * 5 - 17 = 103

x * 5 = 120

x = 24

4 0

3 years ago

Read 2 more answers

Answer the question with explanation;

PSYCHO15rus [73]

Answer:

The statement in the question is wrong. The series actually diverges.

Step-by-step explanation:

We compute

$\lim_{n\to\infty}\frac{n^2}{(n+1)^2}=\lim_{n\to\infty}\left(\frac{n^2}{n^2+2n+1}\cdot\frac{1/n^2}{1/n^2}\right)=\lim_{n\to\infty}\frac1{1+2/n+1/n^2}=\frac1{1+0+0}=1\ne0$

Therefore, by the series divergence test, the series $\sum_{n=1}^\infty\frac{n^2}{(n+1)^2}$ diverges.

EDIT: To VectorFundament120, if $(x_n)_{n\in\mathbb N}$ is a sequence, both $\lim x_n$ and $\lim_{n\to\infty}x_n$ are common notation for its limit. The former is not wrong but I have switched to the latter if that helps.