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1 year ago

6 sin 2 ⁡ x + 4 cos 2 ⁡ x ≡ A + B cos 2 ⁡ x where A and B are integers. Work out the values of A and B .

1 answer:

7 0

$\textit{Pythagorean Identities} \\\\ sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta) \\\\[-0.35em] ~\dotfill\\\\ 6sin^2(x)+4cos^2(x)\implies 6[~1-cos^2(x)~]+4cos^2(x) \\\\\\ 6-6cos^2(x)+4cos^2(x)\implies \stackrel{A}{6}~~ \stackrel{B}{-2}cos^2(x)$

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Theres a pic need asap!!!

miskamm [114]

Answer:

Domain: [-1,3)

Range: (-5,3]

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

A bag contains 5 blue marbles, 2black marbles ,and 3 red marbles. A marine is randomly drawn from the bag The probability of not

notka56 [123]

Answer:

80%; 30%

Step-by-step explanation:

There is 5 blue marbles, 2 black marbles, and 3 red marbles. In other words, there is a total of 10 marbles.

There are only 2 black marbles and 8 marbles that are not black. Therefore, the probability of not drawing a black marble is 8/10 or 4/5 or 80%.

There are only 3 red marbles out of the 10 total marbles. Therefore, the probability of drawing a red marble is 3/10 or 30%.

6 0

2 years ago

Membership to a music club costs $165. Members

dimulka [17.4K]

Answer:

11 music lessons.

Step-by-step explanation:

We know that membership costs $165 and members pay $25 per music lesson.

So, we can write the following expression:

$165+25m$

The 165 represents the one-time membership fee and the 25m represents the cost for m music lessons.

We know that non-members pay no membership fee but their cost per lesson is $40. So:

$40m$

Represents the cost for non-members for m music lessons.

We want to find how many music lessons would have to be taken for the cost to be the same for both members and non-members. So, we can set the expressions equal to each other:

$165+25m=40m$

And solve for m. Let's subtract 25m from both sides:

$165=15m$

Now, divide both sides by 15:

$m=11$

So, at the 11th music lesson, members and non-members will pay the same.

Further Notes:

This means that if a person would only like to take 10 or less lessons, the non-membership is best because there is no initial fee.

However, if a person would like to take 12 or more lessons, than the membership is best because the membership has a lower cost per lesson than the non-membership.

And we're done!

5 0

3 years ago

Help me with this please! :)

Ivenika [448]

Answer:

just do the number +1 and ×4

4 0

2 years ago

Read 2 more answers