The function h is defined by h (x) = x^2 +5. Find h (4a). ????

The diagonals of a rhombus are 8 and 10. The area of the rhombus is 20 40 80

Georgia [21]

RhombusP diagonal is = 8cmQ diagonals = 10cmArea of rhombus = p. q/28 .10/2 =80 / 2 = 40 cm

3 0

4 years ago

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Relative extrema of f(x)=(x+3)/(x-2)

Salsk061 [2.6K]

Answer:

$\displaystyle f(x) = \frac{x + 3}{x - 2}$ has no relative extrema when the domain is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Step-by-step explanation:

Assume that the domain of $\displaystyle f(x) = \frac{x + 3}{x - 2}$ is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Let $f^{\prime}(x)$ and $f^{\prime\prime}(x)$ denote the first and second derivative of this function at $x$ .

Since this domain is an open interval, $x = a$ is a relative extremum of this function if and only if $f^{\prime}(a) = 0$ and $f^{\prime\prime}(a) \ne 0$ .

Hence, if it could be shown that $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ , one could conclude that it is impossible for $\displaystyle f(x) = \frac{x + 3}{x - 2}$ to have any relative extrema over this domain- regardless of the value of $f^{\prime\prime}(x)$ .

$\displaystyle f(x) = \frac{x + 3}{x - 2} = (x + 3) \, (x - 2)^{-1}$ .

Apply the product rule and the power rule to find $f^{\prime}(x)$ .

$\begin{aligned}f^{\prime}(x) &= \frac{d}{dx} \left[ (x + 3) \, (x - 2)^{-1}\right] \\ &= \left(\frac{d}{dx}\, [(x + 3)]\right)\, (x - 2)^{-1} \\ &\quad\quad (x + 3)\, \left(\frac{d}{dx}\, [(x - 2)^{-1}]\right) \\ &= (x - 2)^{-1} \\ &\quad\quad+ (x + 3) \, \left[(-1)\, (x - 2)^{-2}\, \left(\frac{d}{dx}\, [(x - 2)]\right) \right] \\ &= \frac{1}{x - 2} + \frac{-(x+ 3)}{(x - 2)^{2}} \\ &= \frac{(x - 2) - (x + 3)}{(x - 2)^{2}} = \frac{-5}{(x - 2)^{2}}\end{aligned}$ .

In other words, $\displaystyle f^{\prime}(x) = \frac{-5}{(x - 2)^{2}}$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

Since the numerator of this fraction is a non-zero constant, $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ . (To be precise, $f^{\prime}(x) < 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace\!$ .)

Hence, regardless of the value of $f^{\prime\prime}(x)$ , the function $f(x)$ would have no relative extrema over the domain $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

7 0

3 years ago

Which is the graph of x^2/9+y^2/16=1

vodka [1.7K]

Answer:

the answer is A since i can't the letters right is the first one

Step-by-step explanation:

3 0

4 years ago

Xis normally distributed with mean 1,000 and standard deviation 250. What is the probability that X lies between 800 and 1, 100?

Dominik [7]

Answer:

0.4435

Step-by-step explanation:

Given that :

X is normally distributed:

mean(m) = 1,000

standard deviation (s) = 250

probability that X lies between 800 and 1,100?

Using the relation :

X = 800

Zscore = (x - m) / s

Zscore = (800 - 1000) / 250

Zscore = - 200 / 250

Zscore = - 0.8

P(Z ≤ - 0.8) = 0.2119

X = 1100

Zscore = (x - m) / s

Zscore = (1100 - 1000) / 250

Zscore = 100 / 250

Zscore = 0.4

P(Z ≤ 0.4) = 0.6554

P(Z ≤ 0.4) - P(Z ≤ - 0.8)

0.6554 - 0.2119

= 0.4435

5 0

4 years ago

While in college, why did Euler work through advanced math books on his own?

Kay [80]

The correct answer is:

His tutor did not have time to give him private lessons so he was encouraged to study on his own.

Explanation:

Quoting Euler himself in his unpublished, autobiographical writings, his tutor was named Johann Bernoulli.

Euler states "...True, he was very busy and so refused flatly to give me private lessons; but he gave me much more valuable advice to start reading more difficult mathematical books on my own and to study them as diligently as I could; if I came across some obstacle or difficulty, I was given permission to visit him freely every Sunday afternoon and he kindly explained to me everything I could not understand..."

8 0

3 years ago

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