Algebra 2 Problems of the Day (Jan 2020)

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January 2020, Part III

All Questions in Part III are worth 4 credits. Work must be shown. Partial credit is given.


35. Algebraically solve the following system of equations.

(x - 2)² + (y - 3)² = 16
x + y - 1 = 0

Answer:
The first equation is for a circle, and the second is a line. There will be either 1 solution, if it's tangent, or two if it's a secant line. (Zero solutions is unlikely, but not out the question with Regents exams. However, they'd usually be some indication if it was a possibility.)

Solve the linear equation for y:

x + y - 1 = 0
y = -x + 1

Substitute into the first equation:

(x - 2)² + (-x + 1 - 3)² = 16
(x - 2)² + (-x - 2)² = 16
x² - 4x + 4 + x² + 4x + 4 = 16
2x² + 8 = 16
2x² - 8 = 0
x² - 4 = 0
(x + 2)(x - 2) = 0
x = -2 or x = 2

If x = -2, -2 + y - 1 = 0, and y = 3
If x = 2, 2 + y - 1 = 0, and y = -1

Solutions: (-2, 3) and (2, -1)

36. The table below gives air pressures in kPa at selected altitudes above sea level measured in kilometers.

x	Altitude (km)	0	1	2	3	4	5
y	Air Pressure (kPa) (km)	101	90	79	70	62	54

Write an exponential regression equation that models these data rounding all values to the nearest thousandth.
Use this equation to algebraically determine the altitude, to the nearest hundredth of a kilometer, when the air pressure is 29 kPa.

Answer:
First of all "kPa" stands for "kilopascals". It's a unit of pressure, and nothing that you need to concern yourself about, or freak out about if you've never seen it before.

Put the data into the list on the calculator, and run an exponential regression.
y = a(b)^x
a = 101.5228...
b = .8826...
y = 101.523(.883)^x

Use the formula to solve when y = 29.
29 = 101.523(.883)^x
29/101.523 = (.883)^x
0.28564 = (.883)^x
log(0.28564) = x log(.883)
log(0.28564) / log(.883) = x
x = 10.07

Comments and questions welcome.

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