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vagabundo [1.1K]
3 years ago
5

Find the measure to the nearest degree

Mathematics
2 answers:
pantera1 [17]3 years ago
8 0

Answer:

Please see the attached pictures for the full solution.

Sergio039 [100]3 years ago
5 0

Answer:

1. tan A= 1.6643

A = tan^-1(1.6643)

A = 59° , 239°

2. sin A= 0.8480

A = sin^-1(0.8480)

A = 58°, 122°

3. cos A= 0.3420

A = cos^-1(0.342)

A = 1° , 359°

4. cos U= 0.4695​

U = cos^-1(0.4695)

U = 62°, 298°

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Variable g is 5 more than variable w. Variable g is also 2 less than w. Which pair of equations best models the relationship bet
jeyben [28]
Your answer is the first choice, g = w + 5    &    g = w − 2, because if variable g is 5 MORE THAN variable w, that would mean you would have to add 5 to w in order to get g, (g = w + 5), but if it said it was 5 TIMES MORE THAN, THHHHEEEENNNNN that would mean w times 5.
Now da second problem is g = w - 2, because they said that g is 2 LESS THAN w, so you would have to subtract 2 from w to get g, also known as g = w - 2.
I hope I helped! =D
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3 years ago
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Alexeev081 [22]

<u><em>Its 41....VERY EASY</em></u>



Let me show you:

4*4= 16 because of the exponent (2)

4 times 4 =16

Then add by 25..

Which gives u 41..


Hope this helps....


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Answer:

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I found the interval of convergence, but I am not sure how to do the second part, finding the sum of the series as a function of
antiseptic1488 [7]

The given series is geometric with common ratio 6^x - 9, which converges if |6^x - 9| (i.e. the interval of convergence). We have the well-known result

\displaystyle |r| < 1 \implies \sum_{n=0}^\infty ar^n = \frac{a}{1-r}

If you're not familiar with that result, it's easy to reproduce.

Let S_N be the N-th partial sum of the infinite series,

\displaystyle S_N = \sum_{n=0}^N \left(6^x - 9\right)^n = 1 + \left(6^x - 9\right) + \left(6^x - 9\right)^2 + \cdots + \left(6^x - 9\right)^N

Multiply both sides by the ratio.

\left(6^x - 9\right) S_N = \left(6^x - 9\right) + \left(6^x - 9\right)^2 + \left(6^x - 9\right)^3 + \cdots + \left(6^x - 9\right)^{N+1}

Subtract this from S_N to eliminate all the powers of the ratio between 0 and N+1.

\left(1 - \left(6^x - 9\right)\right) S_N = 1 - \left(6^x - 9\right)^{N+1}

Solve for S_N.

S_N = \dfrac{1 - \left(6^x - 9\right)^{N+1}}{10-6^x}

Now as N\to\infty, the exponential term converges to 0 and we're left with

\displaystyle \sum_{n=0}^\infty \left(6^x-9\right)^n = \lim_{N\to\infty} S_N = \boxed{\frac1{10-6^x}}

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2 years ago
Describe what the standard notation of a number look like if the scientific notation of that number has a negative exponent?
yKpoI14uk [10]
4.5 x 10^-5 is just 0.00045 and 1 x 10^-3 is 0.001
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