Rewrite 3x+5y=11 and -5x+7y=9 in slope intercept form

Mathematics

1 answer:

chubhunter [2.5K]2 years ago

4 0

Problem 1

subtract 3 x

5y=-3x+11

divide 5

y=3/5x+2.2'

Problem 2

add 5x

7y=5x+9

divide 7

y=5/7x+1.3'

'notice'

you can put 2.2 into a fraction (11/5) and 1.3 (9/7) if your teacher allows it, but i put them into decimals

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Evaluate the given integral by changing to polar coordinates... \int \int_{D}^{}} xy dA ...where D is the disk with center the o

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$\int\limits_0^{2\pi}\int\limits_0^r r^3cos(\theta)sin(\theta) \, dr \, d\theta = \int\limits_0^{2\pi}(\frac{r^4}{4}cos(\theta)sin(\theta)) \, |_{r = 0}^{r = 7} \, d\theta$

A primitive of cos(Θ)sin(Θ) can be obtained using substitution, and it is sin²(Θ)/2 (note that the derivate of sin²(Θ) is 2sin(Θ)cos(Θ)). Therefore, taking both the dividing 4 and the 2 obtained, we have

$\int\limits_0^{2\pi}(\frac{r^4}{4}cos(\theta)sin(\theta)) \, |_{r = 0}^{r = 7} \, d\theta = \frac{1}{8} \int\limits_0^{2\pi} 7^4 * \frac{cos(\theta)sin(\theta)}{2} \, d\theta = \frac{7^4}{8} (sin^2(\theta)) |_{\theta=0}^{\theta=2\pi} \\= \frac{7^4}{8} (sin^2(2\pi)-sin^2(0)) = 0$