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ExtremeBDS [4]
3 years ago
8

How do I solve? 5 sin 4x = −10 sin 2x

Mathematics
2 answers:
Ulleksa [173]3 years ago
5 0
Move <span><span>5cos<span>(x)</span></span><span>5cos<span>(x)</span></span></span> to the left side of the equation by subtracting it from both sides.<span><span>5sin<span>(2x)</span>−5cos<span>(x)</span>=0</span><span>5sin<span>(2x)</span>-5cos<span>(x)</span>=0</span></span>Simplify each term.Apply the sine double-angle identity.<span><span>5<span>(2sin<span>(x)</span>cos<span>(x)</span>)</span>−5cos<span>(x)</span>=0</span><span>5<span>(2sin<span>(x)</span>cos<span>(x)</span>)</span>-5cos<span>(x)</span>=0</span></span>Multiply <span>22</span> by <span>55</span> to get <span>1010</span>.<span><span>10<span>(sin<span>(x)</span>cos<span>(x)</span>)</span>−5cos<span>(x)</span>=0</span><span>10<span>(sin<span>(x)</span>cos<span>(x)</span>)</span>-5cos<span>(x)</span>=0</span></span>Remove parentheses around <span><span>sin<span>(x)</span>cos<span>(x)</span></span><span>sin<span>(x)</span>cos<span>(x)</span></span></span>.<span><span>10sin<span>(x)</span>cos<span>(x)</span>−5cos<span>(x)</span>=0</span><span>10sin<span>(x)</span>cos<span>(x)</span>-5cos<span>(x)</span>=0</span></span>Factor <span><span>5cos<span>(x)</span></span><span>5cos<span>(x)</span></span></span> out of <span><span>10sin<span>(x)</span>cos<span>(x)</span>−5cos<span>(x)</span></span><span>10sin<span>(x)</span>cos<span>(x)</span>-5cos<span>(x)</span></span></span>.
Factor <span><span>5cos<span>(x)</span></span><span>5cos<span>(x)</span></span></span> out of <span><span>10sin<span>(x)</span>cos<span>(x)</span></span><span>10sin<span>(x)</span>cos<span>(x)</span></span></span>.<span><span>5cos<span>(x)</span><span>(2sin<span>(x)</span>)</span>−5cos<span>(x)</span>=0</span><span>5cos<span>(x)</span><span>(2sin<span>(x)</span>)</span>-5cos<span>(x)</span>=0</span></span>Factor <span><span>5cos<span>(x)</span></span><span>5cos<span>(x)</span></span></span> out of <span><span>−5cos<span>(x)</span></span><span>-5cos<span>(x)</span></span></span>.<span><span>5cos<span>(x)</span><span>(2sin<span>(x)</span>)</span>+5cos<span>(x)</span><span>(−1)</span>=0</span><span>5cos<span>(x)</span><span>(2sin<span>(x)</span>)</span>+5cos<span>(x)</span><span>(-1)</span>=0</span></span>Factor <span><span>5cos<span>(x)</span></span><span>5cos<span>(x)</span></span></span> out of <span><span>5cos<span>(x)</span><span>(2sin<span>(x)</span>)</span>+5cos<span>(x)</span>⋅−1</span><span>5cos<span>(x)</span><span>(2sin<span>(x)</span>)</span>+5cos<span>(x)</span>⋅-1</span></span>.<span>5cos<span>(x)</span><span>(2sin<span>(x)</span>−1)</span>=<span>0</span></span>
Goshia [24]3 years ago
5 0
Answer:

Solve 5sin 4x = -10sin 2x

Ans: 0, pi/2, pi

Explanation:

sin 4x = 2sin 2x.cos 2x
10sin 2x.cos 2x + 10sin 2x = 0
10sin 2x(cos 2x + 1) = 0
a. sin 2x = 0 -->
2x = 0 --> x = 0
2x = pi --> x = pi/2
2x = 2pi x = pi
b. cos 2x = -1 --> 2x = pi --> x = pi/2

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x + y = 53

x -y = 9  Adding both equations

2x = 62

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A^2+b^2+c^4=2020. Yes
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yes

Step-by-step explanation:

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The radius r(t)r(t)r, (, t, )of the base of a cylinder is increasing at a rate of 111 meter per hour and the height h(t)h(t)h, (
sp2606 [1]

Answer:

The volume is decreasing at a rate 20π cubic meter per hour.

Step-by-step explanation:

We are given the following in the question:

The radius is increasing at a rate 1 meter per hour.

\dfrac{dr}{dt} = 1\text{ meter per hour}                

The height is decreasing at a rate 4 meter per hour  

\dfrac{dh}{dt} = -4\text{ meter per hour}

At an instant time t,

r = 5 meter

h = 8 meters

Volume of cylinder =

\pi r^2 h

where r is the radius and h is the height of the cylinder.

Rate of change of volume is given by:

\dfrac{dV}{dt} = \dfrac{d(\pi r^2 h)}{dt}\\\\\dfrac{dV}{dt} = 2\pi rh\dfrac{dr}{dt} + \pi r^2\dfrac{dh}{dt}

Putting all the values we get,

\dfrac{dV}{dt} = 2\pi (5)(8)(1) + \pi (5)^2(-4)\\\\\dfrac{dV}{dt} = 80\pi - 100\pi = -20\pi \approx -62.8

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3 0
3 years ago
. The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to nd
Blababa [14]

The question is:

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation

x²y'' - 7xy' + 16y = 0; y1 = x^4

Answer:

The second solution y2 is

A(x^4)lnx

Step-by-step explanation:

Given the homogeneous differential equation

x²y'' - 7xy' + 16y = 0

And a solution: y1 = x^4

We need to find a second solution y2 using the method of reduction of order.

Let y2 = uy1

=> y2 = ux^4

Since y2 is also a solution to the differential equation, it also satisfies it.

Differentiate y2 twice in succession with respect to x, to obtain y2' and y2'' and substitute the resulting values into the original differential equation.

y2' = u'. x^4 + u. 4x³

y2'' = u''. x^4 + u'. 4x³ + u'. 4x³ + u. 12x²

= u''. x^4 + u'. 8x³ + u. 12x²

Now, using these values in the original equation,

x²(u''. x^4 + u'. 8x³ + u. 12x²) - 7x(u'. x^4 + u. 4x³)+ 16(ux^4) = 0

x^6u'' + 8x^5u' + 12x^4u - 7x^5u' - 28x^4u + 16x^4u = 0

x^6u'' + x^5u' = 0

xu'' = -u'

Let w = u'

Then w' = u''

So

xw' = -w

w'/w = -1/x

Integrating both sides

lnw = -lnx + C

w = Ae^(-lnx) (where A = e^C)

w = A/x

But w = u'

So,

u' = A/x

Integrating this

u = Alnx

Since

y2 = uy1

We have

y2 = (Alnx)x^4 = (Ax^4)lnx

Therefore, the second solution y2 is

A(x^4)lnx

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3 years ago
Tell me the degree of polynomial?
Stella [2.4K]

Answer:

For polynomials in two or more variables, the degree of a term is the sum of the exponents of the variables in the term; the degree (sometimes called the total degree) of the polynomial is again the maximum of the degrees of all terms in the polynomial.

Step-by-step explanation:

hope it helps

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