I'm stuck and don't know what to do

Which of the following expressions have a

Yuki888 [10]

A and D i think.(sry this needed to be 20 characters long)

7 0

3 years ago

There are 45 questions on your math exam. You answered 8/10 of them correctly. How many questions did you answer correctly?

schepotkina [342]

Answer:

36 questions answered correctly

Step-by-step explanation:

8/10 = ?/45

8/10 = 36/45

6 0

3 years ago

Read 2 more answers

Did you know it takes about 40 pounds of olives to make 3 liters of olive oil? Orchard A grew about 2000 pounds of olives. How m

velikii [3]

Answer:

150

Step-by-step explanation:

7 0

3 years ago

Solve for x in the equation x2 + 20x+ 100 = 36.

Rom4ik [11]

Answer:

x=16, x=4

Step-by-step explanation:

2 + 20x+ 100 = 36. $x^{2} + 20x+ 100 = 36\\x^{2} + 20x + 64 = 0\\(x+16)(x+4) = 0\\x =16, x=4$

5 0

3 years ago

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