Howard-Wolowitz
  • Howard-Wolowitz
Check my answers: I"ll medal
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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binkeyboy03
  • binkeyboy03
yes you are correct!!!!!!!!!! good job!
binkeyboy03
  • binkeyboy03
AMAZING WORK!!!
Howard-Wolowitz
  • Howard-Wolowitz
@pitamar

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More answers

binkeyboy03
  • binkeyboy03
dude awsome work
Howard-Wolowitz
  • Howard-Wolowitz
I think you just saying that so I"ll give you a medal
binkeyboy03
  • binkeyboy03
no im not
binkeyboy03
  • binkeyboy03
im being completely honest
Howard-Wolowitz
  • Howard-Wolowitz
@Michele_Laino
Michele_Laino
  • Michele_Laino
answer 3) is correct!
Michele_Laino
  • Michele_Laino
question #2 why you selected option 1?
binkeyboy03
  • binkeyboy03
michele nice catch on #2 didnt see that
Howard-Wolowitz
  • Howard-Wolowitz
that what i thought the answer was
Michele_Laino
  • Michele_Laino
question #1 we have to apply the formulas to your data, so I'm not able to say if your answer is correct! Nevertheless, if you have applied such formulas, I think that your answer is correct!
Michele_Laino
  • Michele_Laino
question #4 the answer is correct!
Michele_Laino
  • Michele_Laino
question #5 I think that your answer is correct, since we can write this: \(\Large y=A \cdot B^x\) so, if we take the logarithm of both sides, we get: \(\Large \log y=\log A+x \log B\)
Howard-Wolowitz
  • Howard-Wolowitz
so the only wrong one is #2 you think
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
sincerely, I don't know, please write the steps you used
Michele_Laino
  • Michele_Laino
please correct me, if I'm wrong: if \(x\) is transformed into \(x^2\), then we have: \((1,5),(9,16),(25,35),(49,67),(81,110)\)
Howard-Wolowitz
  • Howard-Wolowitz
ok so this: The wording of this problem is problematic (which of course is not the fault of the student). When I saw the word "linear," I attempted to find the coefficients a and b for the linear function y = ax + b that would fit the data. (In other words, if I were to take the first data point, (1,5), and substitute x=1 and y=5 into y =ax+b, the equation would have to be true, as it would for each and every one of the other data points.) That failed. So, I decided to use a quadratic (instead of linear) model: y = ax^2 + bx + c. Substituting the first point, (1,5), results in 5 = a*1^2 + b*1 + c. Substituting two more points, one at a time, results in two additional LINEAR equations. These three linear equations can be solved simultaneously to obtain values for a, b and c. I ended up with a = 1, b = 3/2, and c = 5/2. Try substituting these three coefficients into the general form y=a*x^2 + bx + c and verifying for yourself that substitution of any or all of the given points results in an equation that is true.
Michele_Laino
  • Michele_Laino
ok! Nevertheless we have a \(quadratic\; law\) between \(x\) and \(y\) and not linear
Michele_Laino
  • Michele_Laino
we can write this formula: \[\Large y = {\left( {x + \frac{3}{4}} \right)^2} + \frac{{31}}{{16}}\]
Michele_Laino
  • Michele_Laino
so I think you are correct!! Nevertheless we have to make a traslation first, namely: \[\Large x \to x + \frac{3}{4}\]
Howard-Wolowitz
  • Howard-Wolowitz
then im wrong on only two: and two is C you think

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