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Section 1.4 #19: Find an equation of the line that passes through the points (−1,5) and (2,−4).
Solution: We will first find its slope and the use the point-slope form of the line. The slope is
m=−4−52−(−1)=−93=−3.
Using the point-slope form of the equation of the line with slope m=−3 and the point (x1,y1)=(−1,5) we get
y−5=−3(x−(−1))
or
y−5=−3(x+1),
or the slope-intercept form
y=−3x+2.
Section 1.4 #43: Find an equation of the line passing through (3,5) parallel to and perpendicular to the line y=27x+1.
Solution: Since the slope of the given line is 27 the slope of the line parallel is also 27. Therefore using the point-slope form we get the following equation for the parallel line:
y−5=27(x−3),
or in slope-intercept form
y=27x+297.
Since the slope of the given line is 27 the slope of the line parallel is −72. Therefore using the point-slope form we get the following equation for the perpendicular line:
y−5=−72(x−3),
or in slope-intercept form
y=−72x+312.
Section 1.5 #36: Khalid makes an investment at 4% simple interest. At the end of 1 year, the total value of the investment is $1560. How much was originally invested?
Solution: Recall the formula I=Prt. We are told in the problem that t=1 and r=0.04. We seek P but we are not told I. We are told the value of P+I=1560 (note: interest is just the extra that accrues but we are told the total value after 1 year). Therefore in the equation I=Prt we may replace I with 1560−P and substitute our other values to get
1560−P=P(0.04)(1).
Add P to both sides to get
1560=1.04P.
Hence the amount originally invested is
P=15601.04=1500.
Section 1.5 #37: In triangle ABC, angle B is five times as large as angle A. The measure of angle C is 2∘ less than that of angle A. Find the measures of the angles.
Solution: We add the three angles and use the fact that sum of the angles in a triangle is 180 to get
5x+x+(x−2)=180.
Simplify the left-hand isde to get
7x=182,
or and solve for the value of angle A:
x=1827=26.
Therefore angle B is 26⋅5=130 and angle C is 26−2=24.
Section 1.6 #35: Solve and graph the inequality
3x≤−6orx−1>0.
Solution: Divide the left piece 3x≤−6 by 3 to get
x≤−2.
Add 1 to the right piece to get x>1. Now we graph both inequalities simultaneously: