How would I solve this?

Mathematics

1 answer:

Elena-2011 [213]3 years ago

6 0

9514 1404 393

Answer:

not tangent

Step-by-step explanation:

The triangle has side lengths 5, 12, 14. You can compute the "form factor" for the triangle:

a^2 +b^2 -c^2 . . . . . . . . where 'c' is the longest side

= 5^2 +12^2 -14^2

= 25 +144 -196 = -27

The negative value tells you this triangle is obtuse. If AB were a tangent, it would be perpendicular to the radius—the triangle would be a right triangle.

If you compare this calculation to the Pythagorean theorem, you see that the length AB is longer than the length √(25+144) = 13 that is necessary for the triangle to be a right triangle. That means the angle at A is greater than 90°.

<em>Additional comment</em>

This "form factor" calculation is part of the calculation you would do using the Law of Cosines to determine the largest angle. The sign of the "form factor" tells you the sign of the cosine of the angle. Angles whose cosine is negative are greater than 90°. A positive "form factor" indicates an acute triangle.

I find computing the "form factor" in this way makes interpretation of the result fairly easy. For me, it eliminates the confusion I had when the numbers of the Pythagorean theorem didn't add up. For me, it was too much work to figure whether the triangle was acute or obtuse, and I often got it wrong.

For those interested, the angle measure is arccos((a^2+b^2-c^2)/(2ab)). Here, that's arccos(-27/(2·5·12)) ≈ 103.003°.

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Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:

Vanyuwa [196]

Answer:

This series diverges.

Step-by-step explanation:

In order for the series to converge, i.e. $\lim_{n \to \infty} a_n =A$ it must hold that for any small $\epsilon$ >0, there must exist $n_0\in \mathbb{N}$ so that starting from that term of the series all of the following terms satisfy that $|a_n-A|n_0$ .

It is obvious that this cannot hold in our case because we have three sub-series of this observed series. One of them is a constant series with $a_n=1$ , the other is constant with $a_n=3$ , and the third one has terms that are approaching infinity.

Really, we can write this series like this:

$a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}$

If we denote the first series as $b_n=1$ , we will have that $\lim_{k \to \infty} b_k=1$ .

The second series is denoted as $c_k=2^k$ and we have that $\lim_{k \to \infty} c_k=+\infty$ .

The third sub-series $d_k=3$ is a constant series and it holds that $\lim_{k \to \infty} d_k=3$ .

Since those limits of sub-series are different, we can never find such $n_0\\$ so that every next term of the entire series is close to one number.

To make an example, if we observe the first sub-series if follows that A must be equal to 1. But if we chose $\epsilon =1$ , all those terms associated with the third sub-series will be out of this interval $(A-1, A+1)=(0, 2)$ .