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enot [183]
3 years ago
13

NEED HELP NOW WITH EXPLANATION 11=4p-5

Mathematics
1 answer:
fenix001 [56]3 years ago
5 0
Put p by itself by moving 4 and -5 to the other side
11=4p-5, to get rid of the -5, you must do the opposite function which is addition, add 5 to both sides of the equation
11+5=4p-5+5
you’re left with 16=4p
to get rid of 4 you must do the opposite function, 4p means 4*p so you would need to divid both sides by 4
16/4=4p/4
= 4=p
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Express the equation in rectangular form<br> a) r = 2sin(2Θ)<br> b) r = 4cos(3Θ)
Dafna11 [192]

Answer:

a.  sqrt(x^2+y^2) * (x^2+y^2) = 4xy

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r* = 2 sin 2 theta

We know  sin 2theta = 2 sin theta cos theta

r = 2 * 2 sin theta cos theta

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r = 4 y/r * x/r

r = 4xy / r^2

Multiply each side by r^2

r^3 = 4xy

r * r^2 = 4xy

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sqrt(x^2+y^2) * (x^2+y^2) = 4xy

b) r = 4cos(3Θ)

we know that cos 3 theta = cos^3(theta) - 3 sin^2(theta) cos(theta)

r = 4 *cos^3(theta) - 3 sin^2(theta) cos(theta)

Factor out a cos theta

r = 4 *cos(theta){ cos^2 (theta) - 3 sin^2(theta) }

We know that  sin theta = y/r  and cos theta = x/r

r = 4 (x/r) { (x/r)^2 - 3 (y/r)^2}

r = 4 *(x/r) { (x^2/r^2 - 3 y^2/r^2}

Multiply by r

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Multiply by r^2

r^2 *r^2 = 4x { (x^2 - 3 y^2}

r^4 = 4x{ (x^2 - 3 y^2}

We know  r^2 = (x^2+y^2)

(x^2+y^2)^2 = 4x{ (x^2 - 3 y^2}

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3 years ago
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