Rui is a professional deep water free diver.

Mathematics

2 answers:

Vladimir [108]3 years ago

5 0

The correct question is
Rui is a professional deep water free diver. His altitude (in meters relative to sea level), xxx seconds after diving, is modeled by: d(x)=1/2x^2 -10x What is the lowest altitude Rui will reach?

we have that
d(x)=(1/2)x² -10x

we know that
the function is quadratic (a parabola)
so
the lowest altitude (depth) is the vertex

using a graph tool
see the attached figure

the vertex is the point (10,-50)
that means
His altitude (in meters relative to sea level), 10 seconds after diving is 50 meters under the sea level

therefore
the answer is
the lowest altitude is 50 meters under the sea level

Black_prince [1.1K]3 years ago

5 0

Answer:

its -50 meters

Step-by-step explanation:

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(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".