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2 years ago

The equation y = [x - 5] +2 is?

Mathematics

1 answer:

Ymorist [56]2 years ago

5 0

Answer:

Groot

Step-by-step explanation:

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F as a function of x is equal to the square root of quantity 6 x plus 9 , f as a function of x is equal to the square root of qu

Elis [28]

Answer:

A. $\sqrt{6x+9} +\sqrt{6x-9}$

Step-by-step explanation:

We are given the functions f(x) and g(x) as,

$f(x)=\sqrt{6x+9}$

$g(x)=\sqrt{6x-9}$

It is required to find the function ( f+g ) i.e. ( f+g ) (x)

So, $(f+g)(x)=f(x)+g(x)$

i.e. $(f+g)(x)=\sqrt{6x+9}+\sqrt{6x-9}$

Hence, we get that $(f+g)(x)=\sqrt{6x+9}+\sqrt{6x-9}$ .

So, the first option is correct.

4 0

3 years ago

Nevermind, the question has been solved.

san4es73 [151]

okay.... I guess??? lol

7 0

3 years ago

Read 2 more answers

Help asap!!!!<br> Divide x3/4 by X1/6

HACTEHA [7]

It should be 9/2 or 4.5

5 0

2 years ago

F(x) has domain , and range

marta [7]

Domain: {-4, -2, 0, 2, 4}
range: {-2, -1, 0, 1, 2}

7 0

2 years ago

Find the exact value of cos(sin^-1(-5/13))

son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

$\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}$

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}$

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0

2 years ago