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Naily [24]
3 years ago
12

△ABC and △JKL are two triangles such that ∠A≅∠J and ∠B≅∠K. Which of the following would be sufficient to prove that the triangle

s are congruent? A ACJL=BCKL B ∠C≅∠L C AB≅JK D BC≅Jk
Mathematics
1 answer:
Mrrafil [7]3 years ago
5 0
Kanchenjunga are eaywkwk flak an
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2n+5=20?????????????
Oksanka [162]

2n+5=20

Subtract 5 from 5 and 20

2n=15

Divide 15 and 2 by 2

n=7.5

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3 years ago
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How is a ball dropped from a roof interesting ?
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Answer:

because it will bounce high and hit something.

Step-by-step explanation:

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I’m really stuck on this. Would someone please help?
zavuch27 [327]

Since MN forms a 180 degree angle, if you subtract 148.7 from 180, you get 31.3 degrees. This means angles L and N are both 31.3 degrees. Since the parallelogram is 360 degrees total, you subtract 62.6 to account for angles L and N. The remaining degrees unaccounted for are 297.4 degrees which you then divide by two to get the value of angles O and M, which would give you 148.7 degrees for each O and M. This means that angle x is 148.7 degrees

7 0
4 years ago
Akeem is carrying water in a bucket, but is unsure how many gallons it holds. He knows that the bucket and water weigh 32.49 pou
bixtya [17]

Answer:

Step-by-step explanation:

8.3g+2.61=32.49

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5 0
2 years ago
7. Are the ordered pairs below an arithmetic sequence or geometric
DochEvi [55]

Answer:

8. Arithmetic Progression

9. f(9) = 300

Step-by-step explanation:

Given

\{(1,55)\ (2, 45)\ (3, 35)\ (4, 25)\}

Solving (8): Arithmetic or Geometric

We start by checking if it is arithmetic by checking for common difference (d).

d = y_2 - y_1 = y_3 - y_2 = y_4 - y_3

This gives:

d = 45 - 55 = 35 - 45 = 25 - 35

d = -10 = -10 = -10

d=-10

<em>Because the common difference is equal, then it is an arithmetic progression</em>

<em></em>

Solving (8):

f(n) = f(1) + f(n-1)

To find f(9), we substitute 9 for n

f(9) = f(1) + f(9-1)

f(9) = f(1) + f(8)

We need to solve for f(8); substitute 8 for n

f(8) = f(1) + f(8 - 1)

f(8) = f(1) + f(7)

We need to solve for f(7); substitute 7 for n

f(7) = f(1) + f(7 - 1)

f(7) = f(1) + f(6)

We need to solve for f(6); substitute 6 for n

f(6) = f(1) + f(6 - 1)

f(6) = f(1) + f(5)

We need to solve for f(5); substitute 6 for n

f(5) = f(1) + f(5 - 1)

f(5) = f(1) + f(4)

From the function, f(4) = 25 and f(1) = 55.

So:

f(5)=55 + 25

f(5)=80

f(6) = f(1) + f(5)

f(6) = 55 + 80

f(6) = 135

f(7) = f(1) + f(6)

f(7) = 55 + 135

f(7) = 190

f(8) = f(1) + f(7)

f(8) = 55 + 190

f(8) = 245

f(9) = f(1) + f(8)

f(9) = 55 + 245

f(9) = 300

8 0
3 years ago
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