A piecewise function f(x) is defined as shown. f(x) = StartLayout enlarged left-brace 1st Row 1st column negative five-fourths x
+ 90, 2nd column 0 less-than-or-equal-to x less-than 40 2nd row 1st column negative three-eighths x + 75, 2nd column 40 less-than-or-equal-to x less-than-or-equal-to 200 EndLayout Which table could be used to graph a piece of the function? A 2-column table has 3 rows. The first column is labeled x with entries 0, 16, 40. The second column is labeled y with entries 90, 85, 75. A 2-column table has 3 rows. The first column is labeled x with entries 0, 40, 200. The second column is labeled y with entries 90, 40, 0. A 2-column table has 3 rows. The first column is labeled x with entries 40, 120, 200. The second column is labeled y with entries 75, 30, 0. A 2-column table has 3 rows. The first column is labeled x with entries 40, 160, 200. The second column is labeled y with entries 60, 15, 0.
2 answers:
Answer:
A 2-column table has 3 rows. The first column is labeled x with entries 40, 160, 200. The second column is labeled y with entries 60, 15, 0.
Step-by-step explanation:
All of the tables have x=40 in the first column. The function value there is ...
f(x) = (-3/8)(40) +75 = -15 +75 = 60
Only the last table has y = 60 for x = 40.
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A check of the other points in that table shows they are all on the same (second) piece of the graph, as required by the problem statement. Those points are marked with a purple x in the attachment.
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Problem 35
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Problem 37
121/11 = 11
121 = 11*11
If we say 11+11+11+...+11, and have 11 copies of these values added, then we get to a sum of 121
This is probably the easiest way to get the answer assuming repeated values are allowed.
You can stop here if your teacher allows you to use repeated values. If not, then move onto the next section.
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If your teacher requires you to add 11 <u>different</u> numbers, then you can follow this procedure
- Write out eleven copies of "11" in a sequence
- Subtract 2 from the first "11" (to get 9) and add it to the last copy of "11" (to get 13)
- Subtract 4 from the second "11" (to get 7) and add it to the second to last copy of "11" (to get 15)
- Subtract 6 from the third "11" (to get 5) and add it to the third to last copy of "11" (to get 17)
- Subtract 8 from the fourth "11" (to get 3) and add it to the fourth to last copy of "11" (to get 19)
- Finally, subtract 10 from the fifth "11" (to get 1) and add it to the fifth to last copy of "11" (to get 21)
After carefully following those steps, you'll get this sequence (in the exact order shown):
{9, 7, 5, 3, 1, 11, 21, 19, 17, 15, 13}
There are three properties to notice of this sequence
- It's composed of two decreasing arithmetic sequences {9, 7, 5, 3, 1} and {21, 19, 17, 15, 13}
- The 11 in the middle hasn't been changed from the original sequence of nothing but "11"s.
- You should find that the terms of this new sequence add to 121.
That sequence we got sorts to {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21}
and we can say 1+3+5+7+9+11+13+15+17+19+21 = 121
<h3>Answer: 1+3+5+7+9+11+13+15+17+19+21 = 121</h3>