Algebra Know-It-ALL
Then we divide through by 2, getting x=− 3 For Prob. 2 (Table B-2). The standard-form equation we got was 17 x+ 6 = 0 We subtrac ...
626 Worked-Out Solutions to Exercises: Chapters 11 to 19 Multiplying through by 4, we get 2 x+ 8 =− 4 Subtracting 8 from each si ...
Subtracting 20 from each side gives us 3 x= 180 Dividing through by 3, we get x= 60 That means Bonnie weighs 60 kg. Bruce weighs ...
628 Worked-Out Solutions to Exercises: Chapters 11 to 19 This mapping, like the one in the Prob. 2, is not one-to-one, so it ca ...
is not a function. It is not an injection because it’s not one-to-one. That means it cannot be a bijection. The relation does ma ...
630 Worked-Out Solutions to Exercises: Chapters 11 to 19 –6 –4 –2 246 2 4 6 –2 –4 –6 x y (–5,5) (–3,3) (0,0) (2,2) (5,5) Figure ...
–6 –2–4 246 2 4 6 –6 x y (1,0) (5,4) (–3,–4) Figure B-3 Illustration for the solution to Prob. 8 in Chap. 14. –6 –2–4 2 6 2 4 6 ...
632 Worked-Out Solutions to Exercises: Chapters 11 to 19 When the scale increments in a graph are not the same (as is the case ...
Chapter 15 We have these two ordered pairs defining the points P and Q, respectively: P= (u 1 ,v 1 )= (−1,−6) and Q= (u 2 ,v 2 ...
634 Worked-Out Solutions to Exercises: Chapters 11 to 19 We know the coordinates of at least one point (two, actually) and we a ...
angle of −45°. That means the first line will go “uphill” at 45° as we go to the right, and the second line will go “downhill” a ...
636 Worked-Out Solutions to Exercises: Chapters 11 to 19 Dividing through by 2, we get t= 5. That’s the t value of the intersect ...
which simplifies to y− 8 = (x− 2) × (−12) / (−2) and further to y− 8 = 6(x− 2) That’s the PS form of the equation. Once again, ...
638 Worked-Out Solutions to Exercises: Chapters 11 to 19 Let’s get the equations into SI form. In the first equation, we can sub ...
Adding x to each side produces 100 = 7 x Dividing through by 7, we find that x= 100/7. We can plug this into the second original ...
640 Worked-Out Solutions to Exercises: Chapters 11 to 19 Adding 15 to each side gives us −x+ 150 =x Adding x to each side, we ge ...
Let’s state the two equations again for reference, and then try to solve them using double elimination: 2 x+y= 3 and 6 x+ 3 y= ...
642 Worked-Out Solutions to Exercises: Chapters 11 to 19 The slope of this graph is −2, the same as the slope of the graph of th ...
In the first equation, we can subtract x from each side to get y=−x+ 44 We can substitute the quantity (−x+ 44) for y in the sec ...
644 Worked-Out Solutions to Exercises: Chapters 11 to 19 and simplified to x+x+ 83 = 13 When we subtract 83 from each side and n ...
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