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Lorico [155]
2 years ago
12

Can anyone explain those marked steps in case (iii)

Mathematics
1 answer:
Tema [17]2 years ago
5 0
\cos\dfrac{3x}2=\cos\left(\dfrac\pi2+\dfrac x2\right)
\implies \dfrac{3x}2=2n\pi+\dfrac\pi2+\dfrac x2

follows from the fact that the cosine function is 2\pi-periodic, which means \cos x=\cos(2\pi+x). Roughly speaking, this is the same as saying that a point on a circle is the same as the point you get by completing a full revolution around the circle (i.e. add 2\pi to the original point's angle with respect to the horizontal axis).

If you make another complete revolution (so we're effectively adding 4\pi) we get the same result: \cos x=\cos(4\pi+x). This is true for any number of complete revolutions, so that this pattern holds for any even multiple of \pi added to the argument. Therefore \cos x=\cos(2n\pi+x) for any integer n.

Next, because \cos(-x)=\cos x, it follows that \cos x=\cos(2n\pi-x) is also true for any integer n. So we have

\implies \dfrac{3x}2=2n\pi\pm\left(\dfrac\pi2+\dfrac x2\right)

The rest follows from considering either case and solving for x.
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What number is 16% of 44 rounded to the nearest tenth show the work also and thxs
OLEGan [10]
SOLUTION:

16% of 44 = 44 × 16% OR 44 × 16 / 100 OR 44 × 0.16

16% of 44 = 7.04

16% of 44 = 7.0 ( rounded to the nearest tenth )

Hope this helps! :)
Have a lovely day! <3
7 0
3 years ago
Which rational number is greater? <br> —<br> 0.76 or 0.76? And why?
sertanlavr [38]
You have written exactly the same number twice, so neither of them is bigger.
5 0
3 years ago
Given the differential Equation y'+2y=2e^x ;solve this equation using the integration factor; solve for y to get the general sol
SCORPION-xisa [38]

y'+2y=2e^x\Longrightarrow y'=2e^x-2y

If f'(x)=g(x) then y=\int{g(x)dx}

So we extract,

y=\int{2e^x-2x}dx

Which becomes,

y=2e^x-x^2+C

Hope this helps.

r3t40

5 0
3 years ago
Solve the following system of equations: −2x + y = 1 −4x + y = −1
Katyanochek1 [597]

Answer:

<h2>x = 1, y = 3 → (1, 3)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}-2x+y=1\\-4x+y=-1&\text{change the signs}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}-2x+y=1\\4x-y=1\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad2x=2\qquad\text{divide both sides by 2}\\.\qquad x=1\\\\\text{Put it to the first equation:}\\\\-2(1)+y=1\\-2+y=1\qquad\text{add 2 to both sides}\\y=3

5 0
3 years ago
Which system of equations can be used to find the roots of the equation 4x^5-12x^4+6x=5x^3-2x?
Sveta_85 [38]
^{(1)}\ \ 4x^5-12x^4+6x=5x^3-2x\ \ ^{(2)}\\\\  \left\{\begin{array}{ccc}y=4x^5-12x^4+6x\ \ \ ^{(1)}\\y=5x^3-2x\ \ \ ^{(2)}\end{array}\right


Answer: system nr 4.
7 0
3 years ago
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