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2 years ago

What is the range of the function represented by the graph?

Mathematics

2 answers:

mojhsa [17]2 years ago

8 0

Answer:

{1,2,3,4}

Step-by-step explanation:

cant be zero, also got it right on edge 2020

icang [17]2 years ago

7 0

Answer:

{1,2,3,4}

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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2 years ago

Profit 504 thousand more than -141 thousand

polet [3.4K]

The other company's profit was 363 thousand dollars more

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3 years ago

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Lana71 [14]

Answer:

Morgan, its a straight line and it goes through the origin

Step-by-step explanation:

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2 years ago

PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER

Aloiza [94]

Answer:

7 -5i

Step-by-step explanation:

The additive inverse is the number we add to make it equal to zero

-7+5i + x = 0

the real part

-7 + x = 0

x = 7

The imaginary part

5i+x = 0

x = -5i

The complex x is

7 -5i

Notice it is the opposite of the number

- (-7+5i)

7-5i

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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