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1 year ago

Multiply. -8. -9/3 . 2/-5Write your answer in simplest form.

Mathematics

1 answer:

oee [108]1 year ago

3 0

$(-8)\cdot(-\frac{9}{3})\cdot(-\frac{2}{5})$

To multiply these numbers, the first step is to writhe "-8" as an improper fraction, to do so, divide it by 1

$(-\frac{8}{1})\cdot(-\frac{9}{3})\cdot(-\frac{2}{5})$

Next is to solve the multiplication, to do so, first multiply the first two terms of the multiplication:

$(-\frac{8}{1})\cdot(-\frac{9}{3})$

The multiplication is between two negative numbers, when you multiply two negative numbers, the minus signs cancel each other and turn into a positive value, this is called "double-negative"

$(-\frac{8}{1})\cdot(-\frac{9}{3})=\frac{8\cdot9}{1\cdot3}=\frac{72}{3}$

Next multiply the result by the third fraction -2/5

This time you are multiplying a positive and a negative number, so the result of the calculation will be negative

$\frac{72}{3}\cdot(-\frac{2}{5})=-\frac{72\cdot2}{3\cdot5}=-\frac{144}{15}$

Final step is to simplify the result, both 144 and 15 are divisible by 3, so divide the numerator and denominator by 3 to simplify the result to the simplest form:

$-\frac{144\div3}{15\div3}=-\frac{48}{5}$

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3 years ago

(1 point) Find the area enclosed by the curve r=2 sin(0) + 3 sin(90).

olga55 [171]

If the 0's are indeed zeroes, then it would appear that you're using degrees, so that the equation

r(θ) = 2 sin(0°) + 3 sin(90°)

describes a circle r = 3 (since sin(0°) = 0 and sin(90°) = 1). In this case, the area would simply be π r ² = 9π.

But I suspect you meant to use θ, so the curve has the far more interesting equation,

r(θ) = 2 sin(θ) + 3 sin(9θ)

Since sin(9θ) has a period of 2π/9 and sin(θ) has a period of 2π, their sum has a period of 2π. (That is, 2π times the LCM of 1 and 1/9, which is 1.) But we observe r(θ) is odd, since

r(-θ) = 2 sin(-θ) + 3 sin(-9θ)

… = -2 sin(θ) - 3 sin(9θ)

… = - r(θ)

which tells us that the curve closes itself over a half-period. So the area bounded by the curve is given by the integral,

$\displaystyle\int_0^\pi\int_0^{r(\theta)}r\,\mathrm dr\,\mathrm d\theta$

If you're not familiar with double integrals yet, all you need to know is that this reduces to the area formula you may/should be familiar with,

$\displaystyle\frac12\int_0^\pi r(\theta)^2\,\mathrm d\theta$

Compute the integral:

$=\displaystyle\frac12\int_0^\pi(2\sin(\theta)+3\sin(9\theta))^2\,\mathrm d\theta$

$=\displaystyle\frac12\int_0^\pi(4\sin^2(\theta)+12\sin(\theta)\sin(9\theta)+9\sin^2(9\theta))\,\mathrm d\theta$

Apply some identities:

sin²(θ) = (1 - cos(2θ)) / 2

sin(θ) sin(9θ) = (cos(9θ - θ) - cos(9θ + θ)) / 2 = (cos(8θ) - cos(10θ)) / 2

sin²(9θ) = (1 - cos(18θ)) / 2

$=\displaystyle\frac14\int_0^\pi(4(1-\cos(2\theta))+12(\cos(8\theta)-\cos(10\theta))+9(1-\cos(18\theta))\,\mathrm d\theta$

$=\displaystyle\frac14\int_0^\pi(13 - 4\cos(2\theta) + 12\cos(8\theta) - 12\cos(10\theta) - 9\cos(18\theta))\,\mathrm d\theta$

$=\dfrac14 \left(13\theta-2\sin(2\theta)+\dfrac32\sin(8\theta)-\dfrac65\sin(10\theta)-\dfrac12\sin(18\theta)\right)\bigg|_0^\pi$

$=\dfrac1{40} \left(130\theta-20\sin(2\theta)+15\sin(8\theta)-12\sin(10\theta)-5\sin(18\theta)\right)\bigg|_0^\pi$

$=\dfrac{130\pi-20\sin(2\pi)+15\sin(8\pi)-12\sin(10\pi)-5\sin(18\pi)}{40}$

= 13π/4

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3 years ago

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