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ziro4ka [17]
3 years ago
15

Find the measures of two angles, one positive and one negative, that are coterminal with

ac%7B%5Cpi%7D%7B5%7D" id="TexFormula1" title="\frac{\pi}{5}" alt="\frac{\pi}{5}" align="absmiddle" class="latex-formula">.
a. \frac{11\pi}{5} ; \frac{-9\pi}{5} \\b. \frac{\pi}{5} + 360^o ; \frac{\pi}{5} - 360^o \\c. \frac{6\pi}{5} ; \frac{-4\pi}{5} \\d. \frac{11\pi}{5} ; \frac{-\pi}{5}

Mathematics
1 answer:
aleksandrvk [35]3 years ago
5 0

Answer:

a. 11/5 pi; -9/5 pi

Step-by-step explanation:

Coterminal angles are those which have a common terminal side. For example 30° is coterminal with −330° and 390° (see figure).

From the example we can see that the following expressions must be fulfilled:

positive angle - reference angle = 360°

reference angle - negative angle = 360°

where positive angle is 390°, reference angle is 30° and negative angle is -330°. In this problem reference angle is pi/5. Also, we have to change 360° for its equivalent in radians, i. e., 2 pi. So,

positive angle - pi/5 = 2 pi

positive angle = 2 pi + pi/5

positive angle = 11/5 pi

pi/5 - negative angle = 2 pi

negative angle = pi/5 - 2 pi

negative angle = -9/5 pi

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The length of the hypothenuse or the value of x is equal to 19.53

Data;

  • hypothenuse = x
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<h3>Pythagoras's Theorem</h3>

To solve this problem, we have to use Pythagoras's theorem which is used to find the missing side in a right angle triangle if we have at least two sides.

The formula for this is

x^2 = y^2 + z^2\\

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2 years ago
For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
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Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

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