Hey! Can someone please answer this?

2 answers:

7 0

This picture should provide a full answer. What I did was rearrange the equation to get Tan(x) on one side, then in order to get rid of the Tan, I used Arctan, which got us the final answer.

lys-0071 [83]2 years ago

4 0

Answer:

see explanation

Step-by-step explanation:

3tanx - 2 = 4cos35.3° ( add 2 to both sides )

3tanx = 4cos35.3° + 2 ( divide both sides by 3 )

tanx = $\frac{4cos35.3+2}{3}$ ≈ 1.75485

(b)

x = $tan^{-1}$ (1.75485) ≈ 60.3° ( to 1 dec. place )

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PLEASE help me with this sorry if it’s sideways if something is wrong let me know

VladimirAG [237]

Answer: left to right up to down no no yes no yes yes

Step-by-step explanation: hope it helps

6 0

2 years ago

How do find the domain and range of <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7Bx-8%7D%7D" id="TexFormula1" tit

Kisachek [45]

$\sqrt{x-8}$ is undefined if the argument $x-8$ is negative, so you first need to require that

$x-8\ge0\implies x\ge8$

We're not done yet, though, because $\dfrac1{\sqrt{x-8}}$ still doesn't exist when $x=8$ , so we remove this from the domain and we're left with $x>8$ , or in interval notation, $(8,\infty)$

To find the range, consider the limits of the function as you approach either endpoint of the domain.

$\displaystyle\lim_{x\to8^+}\frac1{\sqrt{x-8}}=+\infty$
$\displaystyle\lim_{x\to\infty}\frac1{\sqrt{x-8}}=0$

Since $\dfrac1{\sqrt{x-8}}$ is positive everywhere, the range is $(0,\infty)$