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gogolik [260]
3 years ago
8

Assemble the proof by dragging tiles to the statemens and reasons columns

Mathematics
1 answer:
bonufazy [111]3 years ago
5 0

Answer:

Step-by-step explanation:

                  Statements                              Reasons

1). y║z                                                      1). Given

2). ∠1 ≅ ∠5                                             2). Alternate interior angles

3). ∠3 ≅ ∠6                                             3). Alternate interior angles

4). m∠1 + m∠2 + m∠3 = ∠LAM              4). Angle addition postulate

5). m∠1 + m∠2 + m∠3 = 180°                 5). Def. of m

6). m∠5 + m∠2 + m∠6 = 180°                6). Substitution

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3х2 = 147<br> Solve by undoing
Dmitry_Shevchenko [17]

Answer: +/- 7

Step-by-step explanation:

3x² = 147

To solve for x divide through by three first.

3x² = 147

x² = 49, now we take the square root of both side by trying to apply laws of indices.

√x² = √49

The square root will neutralize the effect of the square because

√a = a¹/² so (x²)¹/², and (x²)¹/² =

x²×¹/² = x, therefore the solution is

x = +/- 7.

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4 years ago
Which of the following could Not be the side length of a right triangle?
ArbitrLikvidat [17]

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Step-by-step explanation:

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8 0
3 years ago
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If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

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3 years ago
Kendell has an account balance of -$127 then she paid $23 for an app on her phone. How much money is in her account?
andre [41]

she will have a balance of negative 150

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