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2 years ago

WILL GIVE BRAINLIEST

Mathematics

1 answer:

myrzilka [38]2 years ago

3 0

Answer:

$\sqrt{11}$

Step-by-step explanation:

The square root of 11 is 3.31... this could be the answer.

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Convert the integral below to polar coordinates and evaluate the integral.

Lelechka [254]

Answer:

$\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}} xy \, dxdy = \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4} r^3 cos(\theta)\sin(\theta) \,\, drd\theta = 16$

Step-by-step explanation:

We are trying to evaluate this integral.

$\int\limits_{0}^{4/\sqrt{2}}\,\,\int\limits_{y}^{\sqrt{16-y^2}} xy \,\,dxdy$

The first thing that we have to do is understand this region in the plane.

$\{ (x,y) \in \mathbb{R} : 0\leq y \leq \frac{4}{\sqrt{2}} \,\, , y \leq x \leq \, \sqrt{16-y^2} \}$

If you graph it looks something like the photo I join.

Now we need to describe that same region in polar coordinates.

That same region in polar coordinates would be

$\{ (r,\theta) : \,\, 0 \leq \theta \leq \frac{\pi}{4} \,\,\, 0\leq r \leq 4 \}$

Now remember that when we do the polar transformation we use the following formula

$\int\limits_{a}^{b} \, \int\limits_{c}^{d} f(x,y) \,dxdy = \int\limits_{\theta_1}^{\theta_2} \, \int\limits_{r_1}^{r_2} r* f(rcos(\theta),rsin(\theta)) \,drd\theta$

Then our integral would be

$\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}} xy \, dxdy = \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4} r^3 cos(\theta)\sin(\theta) \,\, drd\theta = 16$

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3 years ago

Jose traveled 315 miles in 7 hours. What was Jose's unit rate of speed, in miles per hour?

lidiya [134]

Answer:

45 miles in an hour

Step-by-step explanation:

315/7

8 0

3 years ago

Traditionally, the earth's surface has been modeled as a sphere, but the World Geodetic System of 1984 (WGS-84) uses an ellipsoi

sdas [7]

Answer: You are smart, figure it out yourself

Step-by-step explanation:

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2 years ago

Next sequence of 1,-2,-5,-8

melamori03 [73]

Answer:

-11,-14,-17

Step-by-step explanation:

Your numbers are decreasing by 3.

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2 years ago

Solve the simultaneous equation y = x^2 and y = 3x - 2

gayaneshka [121]

Answer:

${x}^{2} = 3x - 2 \\ {x}^{2} - 3x + 2 = 0 \\ (x - 1)(x - 2) = 0 \\ x = 1 \: or \: x = 2 \\ for \: x \: = 1 \\ \: y = {(1)}^{2} = 1 \\ for \: x = 2 \\ y = {(2)}^{2} = 4$

3 0

2 years ago