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3 years ago

What is the sq. root of 7899

Mathematics

1 answer:

Leona [35]3 years ago

5 0

The square root of 7899 is (about) 88.8763186. This can be rounded to 88.9.

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If g(x)=2x-1 and h(x)= swuare root x, find (g o h) (9)

AnnZ [28]

Answer:

Step-by-step explanation:

To evaluate (g ￮ h)(9), evaluate h(9) , then use this value to evaluate g(x)

h(9) = $\sqrt{9}$ = 3, then

g(3) = 2(3) - 1 = 6 - 1 = 5

8 0

3 years ago

Melanie connected a brown garden hose, a green garden hose, and a black garden hose to make one long hose. The brown hose is 10.

natta225 [31]

Answer:

35.65 feet

Step-by-step explanation:

Melanie connected a brown garden hose , a green garden hose and a black garden hose to make one long hose

The brown hose is 10.75 feet

The green hose is 16.4 feet

The black hose is 8.5 feet

Therefore the farthest distance that the one hose can reach can be calculated as follows

= 10.75 + 16.4 + 8.5

= 35.65 feet

Hence the farthest distance that one hose can reach is 35.65 feet

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4 years ago

Which expression is equivalent to 4. 8 + 2. 2w - 1. 4w + 2. 4 ?

Yuri [45]

I don’t know hope it helps

3 0

3 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 $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

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3 years ago

What repeated operation does a negative exponent show?

Zinaida [17]

Answer:

Decreasing value

Step-by-step explanation:

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4 years ago

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