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alekssr [168]
3 years ago
14

Right triangles and trigonometry help!

Mathematics
1 answer:
Maslowich3 years ago
5 0

Step-by-step explanation:

sin \: 45 \degree =  \frac{39}{z}  \\  \\  \therefore \:  \frac{1}{ \sqrt{2} }  =  \frac{39}{z}   \\  \\  \therefore \: z = 39 \sqrt{2}  \\ let \: p \: be \: the \: perpendicular \:  \\  \\\therefore \: tan \: 45 \degree =  \frac{39}{p}  \\  \\ \therefore \:1 = \frac{39}{p}  \\  \\ \therefore \:p = 39 \\  \\ sin \: 60 \degree =  \frac{39}{x}   \\  \\ \therefore \:  \frac{ \sqrt{3} }{2} =  \frac{39}{x}  \\  \\ \therefore \: x =  \frac{78}{ \sqrt{3} }  \\  \\ \therefore \: x =  \frac{78 \sqrt{3} }{ 3 }  \\  \\ \therefore \: x =  26 \sqrt{3} \\\\tan\: 60 \degree =  \frac{39}{y}   \\  \\ \therefore \:  \sqrt{3}  =  \frac{39}{y}  \\  \\ \therefore \: y =  \frac{39}{ \sqrt{3} }  \\  \\ \therefore \: y=  \frac{39 \sqrt{3} }{ 3 }  \\  \\ \therefore \: y =  13 \sqrt{3} \\

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Answer:

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Step-by-step explanation:

Let's start with the first equation; 3x^2 + 3

Substitute 1 (the first digit in a sequence) for x.

3(1^2) + 3

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3(2^2) + 3 Then the second digit.

3(4) + 3

12 + 3 = <u>15</u>

Since the two numbers we have so far are 6 and 15, there is only one sequence this could match. 6, 15, 30, 51.

2(1^2) - 1

2(1) - 1

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This equation represents 1, 7, 17, 31.

These same steps apply to the other equation as well.

1^2 + 2, then 2^2 + 2, then 2^2 + 2, and so on. (But we don't need to do extra work to figure that out.)

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