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4 years ago

To write an equation in function notation, it must be solved for the dependent variable.

Mathematics

1 answer:

Oxana [17]4 years ago

5 0

I think it’s true sorry if I’m wrong but I’m pretty sure it’s true

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Write the equation of the line that passes through the points (-6, -9) and (7, 7).

Rus_ich [418]

Answer:

$\displaystyle y + 9 = \frac{16}{13} ( x + 6 )$

Step-by-step explanation:

The point-slope form of the equation of a line is:

y - k = m ( x - h )

Where m is the slope and (h,k) is a point through which the line passes.

Given a line passes through points A(x1,y1) and B(x2,y2), the slope can be calculated with the equation:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

We are given the points (-6,-9) and (7,7), thus the slope is:

$\displaystyle m=\frac{7+9}{7+6}=\frac{16}{13}$

Now we apply the point-slope formula using the point (-6,-9):

$\mathbf{\displaystyle y + 9 = \frac{16}{13} ( x + 6 )}$

This equation is expressed in the required format

6 0

3 years ago

Solve for v:<br> -5.5 = 7.3 + v/8

Romashka [77]

-5.5=7.3+v/8

convert decimals to fractions

-11/2=73/10+v/8

multiply both sides by 40

-220=292+5v

move terms

-5v=292+220

add the numbers

-5v=512

divide both sides by -5

v=512/5

answer

3 0

3 years ago

pls help me with this question. just one! thank you so much to whoever decides to it will make my day a lot ❤️❤️

Yuliya22 [10]

Answer:

2 385/3913

i think..

Step-by-step explanation:

6 0

3 years ago

Please select the best answer thanks

alukav5142 [94]

The first one.............

4 0

3 years ago

Read 2 more answers

Evaluate the cosine if the angle of rotation which contains the point (9, -3) on its terminal side

Liono4ka [1.6K]

so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant

$\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{9}}{\stackrel{hypotenuse}{\sqrt{90}}}\implies \stackrel{\textit{rationalizing the denominator}}{\cfrac{9}{\sqrt{90}}\cdot \cfrac{\sqrt{90}}{\sqrt{90}}\implies \cfrac{9\sqrt{90}}{90}}\implies \cfrac{\sqrt{90}}{10}\implies \cfrac{3\sqrt{10}}{10}$

6 0

3 years ago