Help! Pre-Calculus Homework, solving inequalities

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saxonchick

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Here's the question:
The number D (in thousands) of earned degrees conferred annually in the Unites States from 1950 to 1995 is approximated by the model

D = 0.0977t2 + 47.117t + 291.5651

where t = 0 represents 1950.
a) Use a graphing utility to graph the model.
b) According to this model, estimate when the number of degrees will exceed 2,500,000.

I've looked through my book and everything, and can't seem to figure it out.
Will someone please explain this to me?
Thanks!

-Kinsey
 
saxonchick said:
… D = 0.0977t^2 + 47.117t + 291.5651 Please use the caret symbol ^ to show exponents.

where t = 0 represents the number of years since 1950

a) Use a graphing utility to graph the model …

… I've looked through my book and everything, and can't seem to figure it out …
Click to expand...


Hello Kinsey:

I understand from the instructions that you're supposed to use a graphing utility, but is the book you're talking about the user's guide for your calculator?

What is it about graphing this quadratic polynomial that you cannot figure out?

What model calculator are you using?

Do you realize that you might have to use the symbol x instead of t when entering the equation to graph? (This depends upon the calculator.)

In an (x, y) coordinate system, D is y, of course. Most Texas Instruments graphing calculators use the notation Y1 for y.

In addition to graphing the given model, we could also plot the line y = 2500 on the same graph. Then, it's simply a matter of finding the point of intersection between the two graphs. Values of x for which the parabola lies above the horizontal line comprise the solution set for t in part (b).

Many graphing calculators have built-in functions for finding intersections points; they're available after you have the graphs displayed. Otherwise, you can zoom and trace (repeated as much as necessary), to find the intersection point.

Please reply with the steps that you've tried, so that we might determine why you're stuck.

Cheers,

~ Mark 8-)

MY EDIT: Corrected numerical error in the equation y = 2500.
 
What I can't understand is how to solve it, how to find the answers. It's been awhile since I've used a graphing calculator, so I'm having trouble using it and I just want someone to explain to me EXACTLY how to solve this problem, because the book explanation is horrid.
And it is my Pre-Calculus book, not a instruction book.
Please Help! It's due tomorrow and I usually don't have this much trouble with math, for some reason I'm really having issues with this one.
 
saxonchick said:
… for some reason I'm really having issues with this one.
Click to expand...


Exactly. There is some reason why you're stuck, and nobody here will know what these issue are until you tell us.

There is nothing to solve, for part (a).

What model calculator are you using?

If you want to solve part (b) by hand (instead of graphically by using the calculator), then you could start by realizing the following.

D represents the number of degrees in thousands. So, D takes on the value 2,500, when we're talking about 2,500,000 degrees awarded.

Therefore, substitute 2500 for D in the given model, and use the Quadratic Formula to solve the equation for t.

0.0977t^2 + 47.117t + 291.5651 = 2500

Since the variable t represents a number of elapsed years, it must be positive. Therefore, ignore the negative solution for t.

The positive solution represents the number of years after 1950 when the number of degrees awarded is 2,500,000. The approximation required for part (b) is achieved by rounding off; they didn't specify any particular precision, so I think that rounding to the nearest whole number is good enough.

Add that whole number of years to 1950, to report the approximate year in which the number of awarded degrees exceed 2,500,000.

I don't know if you're trying to use the calculator versus paper & pencil. (The instructions that you posted state calculator use; most precalculus texts do not include specific calculator lessons.)

If you need more help, please reply with the steps that you've tried, so that we might determine why you're stuck. 8-)

 
What's with the graph?

\(\displaystyle Look, \ D(t) \ = \ 97.7t^{2} \ + \ 47,117t \ + \ 291,565.1\)

\(\displaystyle D(0) \ = \ 291,565.1 \ which \ means \ in \ 1950 \ (t=0), \ 291,565.1 \ degrees \ were \ awarded .\)

\(\displaystyle Hence, \ 2,500,000 \ = \ 97.7t^{2} \ + \ 47,117t \ + \ 291,565.1, \ t \ equals \ about \ 43 \ years.\)

\(\displaystyle Ergo, \ if \ the \ equation \ holds, \ then \ after \ 1993 \ (1950+43) \ more \ than \ 2,500,000 \ degrees \ will \ be\)

\(\displaystyle \ awarded \ annually.\)

Note: Watch your units as they will try to trick you on them.
 
I made a HUGE mistake when I posted the problem here, it's actually .0977t^2 + 47.1174t >291.5651

How much does this affect the answer?

And thanks so much for your help! You guys have been extremely useful, unlike other experiences I've had at other math boards.

And I just need help with part b, not part a.
 
saxonchick said:
… it's actually .0977t^2 + 47.1174t >291.5651

How much does this affect the answer …
Click to expand...


Hmmm. It seems to be a different question.

Is the following correct?

The model of degrees awarded (D) in terms of the number of years after 1950 (t) is given by:

0.0977t^2 + 47.1174t = D

And now you're asked to solve the following inequality.

0.0977t^2 + 47.1174t > 291.5651

Solving this inequality tells us (approximately) how many years it took for the awards to exceed 291 degrees.

(Why they would choose 291.5651 as some number of awards is somewhat strange, to me.)

In other words, the value 2,500,000 is no longer a part of the exercise, and D is not given in thousands.

If so, then treat the inequality as an equation, and move all terms to one side.

0.0977t^2 + 47.1174t - 291.5651 = 0

Finish by following the same instructions that I posted above regarding the Quadratic Formula.

My gut feeling is that I still do not understand the actual exercise. I think your original version makes more sense.

~ Mark

 
\(\displaystyle Now \ you \ are \ giving \ us \ an \ inequality.\)

\(\displaystyle For \ the \ inequality \ .0977t^{2}+47.1174t \ > \ 291.5651 \ to \ ring \ true, \ t \ must \ be \ in \ the \ interval \\)

\(\displaystyle (-\infty,-488.38)U(6.11,\infty).\)

\(\displaystyle Now, \ what \ has \ this \ to \ do \ with \ your \ first \ thread?\)
 
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