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saveliy_v [14]
3 years ago
9

Max's trip home takes 3 2 32 minutes. What is the latest time he can leave to be home by a quarter before 5 5?

Mathematics
1 answer:
Phantasy [73]3 years ago
3 0

Answer:

Latest time he can leave to be home by a quarter before 5 is 4:13

Step-by-step explanation:

Given Max's trip home takes 32 minutes. we have to find the time at which he can leave to be home by a quarter before 5.

quarter before 5 means 4:45

Max's takes 32 min to come to home so he has to leave 32 minutes before the given time.

Hence, latest time he can leave to be home by a quarter before 5 is 4:45-32 = 4:13

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Step-by-step explanation:

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Answer:

  • 39°
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Step-by-step explanation:

1.

Hypotenuse = 32\\Adjacent = 25\\Angle\:\alpha \:=?\\Using \: SOHCAHTOA ;\\Cos \:\alpha = \frac{adj}{hyp} \\Cos \:\alpha = \frac{25}{32}\\\\\alpha = Cos^-^1\frac{25}{32} \\\alpha =38.62\\\\\alpha = 39

2.

Opposite = 8\\Adjacent = 16\\\alpha =?\\Using SOHCAHTOA\\Tan\:\alpha = \frac{opp}{adj} \\Tan\:\alpha = \frac{8}{16} \\Tan \:\alpha =1/2\\\alpha = Tan^-^11/2\\\\\alpha =26.56\\\alpha =27

3.

Hypotenuse = 41\\Opposite = 19\\\alpha = ?\\Using\: SOHCAHTOA\\Sin\:\alpha = \frac{opp}{hyp} \\Sin \:\alpha =\frac{19}{41}\\\\\alpha =Sin^-^1 (19/41)\\\\\alpha =27.6\\ \alpha =28

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Step-by-step explanation:

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Linear Algebra: Permutation Matrices Let M be the matrix {{0,0,0,1,0}, {0,0,1,0,0}, {0,1,0,0,0}, {0,0,0,0,1}{1,0,0,0,0}}. What i
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Multiplying M by any matrix A would return new matrix, B, in which

• the 1st row of B is equal to the 4th row of A,

• the 2nd row of B is equal to the 3rd row of A,

• the 3rd row of B is equal to the 2nd row of A,

• the 4th row of B is equal to the 5th row of A, and

• the 5th row of B is equal to the 1st row of A.

The pattern here is

1 => 4 => 5 => 1

2 => 3 => 2

Let {4, 3, 2, 5, 1} denote the matrix M, where each number refers to the row of the identity matrix, I.

Using this notation, the pattern above gives

M² = {5, 2, 3, 1, 4}

M³ = {1, 3, 2, 4, 5}

M⁴ = {4, 2, 3, 5, 1}

M⁵ = {5, 3, 2, 1, 4}

M⁶ = {1, 2, 3, 4, 5}

so that <em>n</em> = 6.

(Notice that the first cycle has length 3 and the second one has length 2; the minimum <em>n</em> needed here is then LCM(2, 3) = 6.)

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Step-by-step explanation:

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