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3 years ago

Evaluate the double integral i=∫∫dxyda where d is the triangular region with vertices (0,0),(4,0),(0,4).

1 answer:

8 0

$I=\displaystyle\iint_D\mathrm dA=\int_{x=0}^{x=4}\int_{y=0}^{y=4-x}\mathrm dy\,\mathrm dx$
$I=\displaystyle\int_{x=0}^{x=4}(4-x)\,\mathrm dx$
$I=8$

To verify this, we can simply find the area of the triangle using the well-known formula $\dfrac12bh$ , where $b$ and $h$ are the base and height, respectively of the triangular region $D$ . We have $b=h=4$ , so the area is $\dfrac{4^2}2=8$ , as expected.

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A number is chosen at random from the integers 10 to 30 inclusive. Find the probability that the number is a multiple of 3.

kirill [66]

Answer:

For each part, find how many numbers have the stated characteristic, then divide by 21.

e.g. multiples of 5:

these would be 10, 15, 20, 25, and 30

so prob (a multiple of 5 ) = 5/21

Do the others the same way

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2 years ago

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Please help! Worth 36 points!

CaHeK987 [17]

Answer:

a. Same Side Interior Angles

b. Corresponding Angle

c. Alternate Exterior Angle

d. Alternate Interior Angles

e. 20

f. 70

Step-by-step explanation:

e. Corresponding angle add up to 90 degrees. Which you have to subtract 110 to 90 and is 20

f. same side interior angle adds up to 180 degrees. Which you have to subtract 180 to 110 and it is 70

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2 years ago

Tanya purchased 4 hammers for $11.79 each and 7 screwdrivers for $6.65 each. Write an expression for the total cost of the tools

igor_vitrenko [27]

<span>Write an expression for the total cost of the tools.
</span>
total cost = <span>$11.79 (4) + </span>$6.65 (7)

<span>Then find the total cost.

</span>total cost = $11.79 (4) + $6.65 (7)
total cost = $47.16 + $46.55 = $93.71

7 0

3 years ago

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3. 18 is 40% of what number?

Mariulka [41]

#Lets just write x as the nunber you asked

x × 40/100 = 18

x = 18 : 40/100

x = 18 × 100/40

y = 1800/40

y = 45

<h3>Answer : 45</h3>

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2 years ago

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Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago