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Doss [256]
3 years ago
9

Select the correct answer for each statement. (k+2, k+3 , k+4) or (k+10, k+12, k+14) will yield consecutive odd integers. (k+6,

k+8 , k+10) or (k+1, k+2, k+3) will yield consecutive integers.
Mathematics
2 answers:
Dahasolnce [82]3 years ago
7 0

2k+9, 2k+11, 2k+13 will yield consecutive odd integers.

k+6, k+8, k+10 will yield consecutive integers.

almond37 [142]3 years ago
4 0

Answer:

Step-by-step explanation:

one

The only really certain way of getting an odd integer is to do 2k + 1. Therefore the answer should be something like 2k+1, 2k+3, 2k + 5

The best I can do for the first one is to assume that k is odd and that the answer is k + 10, k+12, k+14, but I would ask your instructor if this is merely the best answer out of a poor lot or if the question really has no answer.

Two things should be noted.

1. k has to be odd.

2. The first choice has to give consecutive integers because what is added on is 2 3 and 4. Those numbers are consecutive.

Two

k+1, k+2, k+3. See point 2 above. k is a constant and any number. 1,2,3 are consecutive so the results are consecutive. Even if k < 0 the numbers will be consecutive.

k= - 10

k+1 = - 9

k+2 = -8

k+3 = - 7

These are consecutive.

Your first choice will give either consecutive odd or even numbers depending on what k is.

If k = even, then k+6, k+ 8, k+10 will all be even.

If k = odd then the givens will be odd.

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I need help on 25,27, 28,29,31 100 points for who answers it right
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Answer:

25. D          27. D       28. A       29. D. x = 7, y = 26       31.  D. 120 degrees

Step-by-step explanation:

25. Since the two parallelograms are congruent, you know what ∠B and ∠Y are supplementary.

This means that 3m + 70 + 80 = 180

3m + 150 = 180

3m = 30

D. m = 10

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E is 5 up and 5 over, so it can still count as a possibility.

28. When you observe the coordinates, you'll notice that FG ║ HI, and FH ║ GI. This eliminates the possibility of it being a kite and ensures that it will at least be a parallelogram!

From here, you can also observe that FG ≅ HI, and FH ≅ GI. This shows you that it's not a square despite that it has right angles.

The presence of right angles shows that it's not a rhombus and instead must be a rectangle.

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6(7) - 8

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