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Applied Finite Mathematics HW Solutions 10

# Applied Finite Mathematics HW Solutions 10 - 9 10 11 12...

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Unformatted text preview: 9. 10. 11. 12. 13‘ . 14. 15. 8 _ EXERCISES 1.3 To ﬁnd the m—intercept of the line, we set 3; = U in the equation of the line. This gives :c. = 7. To ﬁnd the y-intercept of the line, we set :1: = D in the equatiOn of the line. This gwes ——2y 2 7. from which 3; 2: ~7f2. The line is showu in the left ﬁgure below. (1.213) The ﬁne passes through the origin. When we set 3; = 1, We obtain 2 — 33; = o, from which 3; = 2/3. A second point on the line is (1.2/3). The line is shown in the right ﬁgure above. This is a vertical line 4 units to the right of the y—axis (left ﬁgure below). This is a horizontal line 2 units below the m—axis (right ﬁgure shove). 4 + 3 — —Z. The equation of the line is (a) The slope of the ﬁne is m = _2 _ 1 “ 3 7 y+3=*§(93—1) => 3y+9=~7e+7 => 7\$+3y=-—2. ( h) 1When we substitute :1: = 3 into the equation of the line, we get 7(3)+3y=—2 => 3y=—2—21=~23'=> y=__, The point is (3. —23/3). (3.) By writing the given equation in the form y = gs: — 2, we see that the slope of this line is 3 f 2. This must also be the slope of the required line. The equation of the required line is therefore y+1=g(e:—-l) => 2y+2=3\$—3 => 3m—2y=5- (h) To ﬁnd the w-intercept of the line, we set y = D in the equation of the line. We get 33:5 => 59:2. 1 . (a) By writing the given equation in the form 3; = ~33: + 2, we see that the slope of this line is —1/3. The slope of the required line must therefore be its negatiire reciprocal 3. The equation of the required line is therefore 9+4=3(§c+3) => y+4=3e+9 =>3e—y=—5. (h) To ﬁnd the y-interoept of the line, we set a: = D in the equation of the line. We get This material has been reproduced in accordance with oopjm'glﬁ law byﬂae Universitv ofManiloha Emma-e puree» .enma.m:m. :.. --- -- =_ n-n 2- . - - (/F‘Hx ...
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