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3 years ago

Which function has the greatest rate of change?

2 answers:

4 0

For an equation expressed as Ax + By + C, the rate of change is equal to -A/B. For the first equation, it is 6/7. For the second equation, the slope is 2. Then for the third equation, m = (19 - 7) / (4 -2) = 6. Thus, the greatest slope is in third choice.

Ilia_Sergeevich [38]3 years ago

3 0

The rate of change is the slope...and in y = mx + b form, the slope is in the m position. So we need to put the equations in y = mx + b form.

A. 6x - 7y = 21
-7y = -6x + 21
y = 6/7x - 2....so the slope here is 6/7 or 0.857

B. y = 2x - 2....this is already in the form we need...the slope is 2

C. (2,7)(3,13)
slope = (13-7) /(3-2) = 6/1 or just 6

the slope with the greatest rate of change is gonna be the one with the highest slope......and that would be C.

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2 years ago

Calculating cos-1 ( help is gladly appreciated :) )

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Answer:

$\frac{3\pi}{4}$

(Assuming you want your answer in radians)

If you want the answer in degrees just multiply your answer in radians by $\frac{180^\circ}{\pi}$ giving you:

$\frac{3\pi}{4} \cdot \frac{180^\circ}{\pi}=\frac{3(180)}{4}=135^{\circ}$ .

We can do this since $\pi \text{ rad }=180^\circ$ (half the circumference of the unit circle is equivalent to 180 degree rotation).

Step-by-step explanation:

$\cos^{-1}(x)$ is going to output an angle measurement in $[0,\pi]$ .

So we are looking to solve the following equation in that interval:

$\cos(x)=-\frac{\sqrt{2}}{2}$ .

This happens in the second quadrant on the given interval.

The solution to the equation is $\frac{3\pi}{4}$ .

So we are saying that $\cos(\frac{3\pi}{4})=\frac{-\sqrt{2}}{2}$ implies $\cos^{-1}(\frac{-\sqrt{2}}{2})=\frac{3\pi}{4}$ since $\frac{3\pi}{4} \in [0,\pi]$ .

Answer is $\frac{3\pi}{4}$ .

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