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kumpel [21]
3 years ago
14

Can you Solve for c?​

Mathematics
1 answer:
Gennadij [26K]3 years ago
3 0

Answer:

3\sqrt{2} or \sqrt{18}

Step-by-step explanation:

Use 45-45-90 triangle theorem

3*\sqrt{2}

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Find the maximum rate of change of f(x; y) = 4y p x at the point (4,1) and the direction in which it occur
olga2289 [7]

Answer:

16p

Direction: positive/upward direction

Step-by-step explanation:

First, the function is f(x,y) = 4ypx

At the point where x=4 and y=1, the maximum rate of change of the function f(x,y) is

4(1)p(4) = 16p

The direction in which it occurs is the positive(upward) direction

5 0
3 years ago
Please anyone help bend it ASAP
sergejj [24]
Here are the answers:
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10. -72<-70
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4 0
3 years ago
(98 pts + Brainliest) Question regarding logarithims and point slope form
bezimeni [28]

You are right in thinking that the base of the logarithm doesn't matter. It only affects the spread of the data points if you were to plot them, but would not ultimately have any effect on the slope of the line (but it would on the y-intercept).

One major discrepancy I'm noticing is in the values you found for \ln\sin\theta. For example, if \sin\theta=0.05, then you should have \ln\sin\theta=\ln0.05\approx=-3.0. Not sure how you got -7.0, and the same goes for the rest of your table of values.

Another thing is that the provided solution suggests you take the average the first and last pairs of consecutive data points, and use these values in the slope formula to obtain the best-fit line's slope. If that's the case, then you should have

\dfrac{\Delta\ln x}{\Delta\ln\sin\theta}=\dfrac{\frac{1.86+1.72}2-\frac{2.65+2.28}2}{\frac{-5.4-5.3}2-\frac{-7.0-6.4}2}=\dfrac{1.79-2.47}{-5.35+6.7}=\dfrac{-0.68}{1.35}\approx-0.5

(i.e. you have to take the average of the given values, then use those averages in the \Delta expressions - but this doesn't significantly affect the slope you found)

Ultimately, I think the problem is that your expression for the slope appears to be \dfrac{\Delta\ln x}{\Delta\ln\sin\theta}, when the solution says it should be the reciprocal. I'm of the opinion that your slope is correct, since the experiment refers to \theta (and hence \sin\theta) as the independent variable, and so \Delta\ln\sin\theta would serve as the "run" and \Delta\ln x would serve as the "rise".

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3 years ago
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