I need help with this

Mathematics

1 answer:

pochemuha3 years ago

7 0

Answer:

$A(x) = \frac{ 4x + 800 }{x}$

Step-by-step explanation:

$C(x) = 4x + 800.... (given)$

$\because \: A(x) = \frac{C(x) }{x}$

$\therefore \: A(x) = \frac{ 4x + 800 }{x}$

You might be interested in

Which equation is equivalent to the formula m = ca? a = cm a equals m over c c = am c equals a over m.

disa [49]

The option b which is $a=\dfrac{m}{c}$ is correct answer.

Given to us:

$m=ca$

dividing both sides by c, we get

$m=ca\\\\\dfrac{m}{c}=\dfrac{ca}{c}$

canceling c from right hand side, we get

$\dfrac{m}{c}=a \\\\a=\dfrac{m}{c}$

Therefore, the option b which is $a=\dfrac{m}{c}$ is correct answer.

To know more visit:

brainly.com/question/9559346

8 0

2 years ago

Read 2 more answers

Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val

zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

$\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill$

$\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}$

$\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}$