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lapo4ka [179]
4 years ago
7

What is the second term of the sequence generated by the formula an = 3n?

Mathematics
2 answers:
Drupady [299]4 years ago
8 0

Answer:

Option D is correct that is 6 is 2nd term of the sequence.

Step-by-step explanation:

We are given with a formula which generates the terms of a sequence,

a_n=3\times n

we have to find 2nd term,

For the second term we put value of n = 2

Thus,

a_2=3\times2

a_2=6

Therefore, Option D is correct that is 6 is 2nd term of the sequence.

astraxan [27]4 years ago
6 0
I think b.3 i am not sure but it should be b
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Answer:

You have a thing on there saying the answer is c. If you got the answer there is no point of asking it.

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Verify that:
Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)

Note that cot(x) = cos(x) / sin(x). Hence:

\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)

Multiply:

\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Split:

\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Simplify:

\csc(x)-\sin(x)=\csc(x)-\sin(x)

Problem 2)

We want to verify that:

\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}

Square:

\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

\displaystyle \frac{1}{\sin^2(x)}-\frac{2\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}

Factor out the sin²(x) from the denominator:

\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}

Factor (perfect square trinomial):

\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}

Factor (difference of two squares):

\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Factor out a negative from the first factor in the denominator:

\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Cancel:

\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Distribute the negative into the numerator. Therefore:

\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}

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Arturiano [62]

Answer:

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Step-by-step explanation:

<u>The graph is shown in attached image.</u>

<u />

The black line is Juan's graph.

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The end of the curve(s) means the end of the race. Both curve's ending point in y-axis is 4 miles so the end of the race is 4 miles.

But in x-axis, we see the time:

Juan finishes at 13 minutes

Antonio finishes at 15 minutes

<u>Definitely Juan wins the race</u>

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3 years ago
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