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Andrews [41]
3 years ago
12

Determine two pairs of polar coordinates for the point (5, 5) with 0Á _ _ < 360Á

Mathematics
1 answer:
Natasha2012 [34]3 years ago
8 0
The polar coordinates are given by  M(x, y), such tha x=rcos(teta), and y=rsin(teta), it was given M(5,5), so 5=rcos(teta) and 5=rsin(teta), the ratio gives us 5/5= cotan (teta), 1= cotan (teta), implies 1= tan(teta), so teta = Pi/4, let's find r
5=rsin(Pi/4)=r . sqrt(2)/2, so r =10/sqrt(2)=10sqrt(2)
finally the polar coordinates are (r, teta) = (10sqrt(2), Pi/4)
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Find the indefinite integral ∫((x^2)/(4x^3+9)) dx
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Answer:

\displaystyle  \frac{1}{12}   \ln( |{4x}^{ 3}  + 9|)  +  \rm C

Step-by-step explanation:

we would like to integrate the following indefinite integral:

\displaystyle \int  \frac{ {x}^{2} }{4 {x}^{3}  + 9} dx

in order to integrate it we can consider using u-substitution also known as the reverse chain rule and integration by substitution as well

we know that we can use u-substitution if the integral is in the following form

\displaystyle \int f(g(x))g'(x)dx

since our Integral is very close to the form we can use it

let our u and du be 4x³+9 and 12x²dx so that we can transform the Integral

as we don't have 12x² we need a little bit rearrangement

multiply both Integral and integrand by 1/12 and 12:

\displaystyle  \frac{1}{12} \int  \frac{ 12{x}^{2} }{4 {x}^{3}  + 9} dx

apply substitution:

\displaystyle  \frac{1}{12} \int  \frac{ 1}{u} du

recall Integration rule:

\displaystyle  \frac{1}{12}   \ln(|u|)

back-substitute:

\displaystyle  \frac{1}{12}   \ln( |{4x}^{ 3}  + 9|)

finally we of course have to add constant of integration:

\displaystyle  \frac{1}{12}   \ln( |{4x}^{ 3}  + 9|)  +  \rm C

and we are done!

6 0
3 years ago
Find the area of a parallelogram with bass 15 yards and height 21 2/3yards
r-ruslan [8.4K]
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</span><span>5⋅65

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Answer:
325
4 0
3 years ago
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mars1129 [50]

Answer:

It’s 372

Step-by-step explanation:

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4 years ago
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6 0
3 years ago
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tekilochka [14]
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4 years ago
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