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ANTONII [103]
2 years ago
15

Write the Prime factors that 24 and 36 have in common By showing them multiplied together

Mathematics
1 answer:
lakkis [162]2 years ago
5 0

The prime factors that the numbers 24 and 36 have in common are found out to be; 2 × 2 × 3

<h3>How to get prime factors? </h3>

A prime factor of a number is simply factor of that number that is a prime number. This means that any of the prime numbers that can be multiplied to give the original number.

Now, we have;

Prime factors of 36; 2 × 2 × 3 × 3

Prime factors of 24; 2 × 2 × 2 × 3

The prime factors that they have in common will be; 2 × 2 × 3

Read more about Prime factors at; brainly.com/question/1081523

#SPJ1

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Then the integral becomes

\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv

where R' is the unit circle,

\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1

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\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}

Then

\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}

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