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3 years ago

How big is 2000 sq.ft in yards?

Mathematics

2 answers:

melamori03 [73]3 years ago

7 0

200 square feet = 222.2222 square yards.

topjm [15]3 years ago

6 0

There are three feet in a yard and nine square feet in a square yard. Divide 2000 by 3 and then divide by 3 again to get 222.222 square yards.

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Let C be the curve of intersection of the parabolic cylinder x^2 = 2y, and the surface 3z = xy. Find the exact length of C from

Maslowich

I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).

The arc length can be computed with a line integral, but first we'll need a parameterization for $C$ . This is easy enough to do. First fix any one variable. For convenience, choose $x$ .

Now, $x^2=2y\implies y=\dfrac{x^2}2$ , and $3z=xy\implies z=\dfrac{x^3}6$ . The intersection is thus parameterized by the vector-valued function

$\mathbf r(x)=\left\langle x,\dfrac{x^2}2,\dfrac{x^3}6\right\rangle$

where $0\le x\le 4$ . The arc length is computed with the integral

$\displaystyle\int_C\mathrm dS=\int_0^4\|\mathbf r'(x)\|\,\mathrm dx=\int_0^4\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}\,\mathrm dx$

Some rewriting:

$\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}=\sqrt{\dfrac{x^2}{36}}\sqrt{x^4+9x^2+36}=\dfrac x6\sqrt{x^4+9x^2+36}$

Complete the square to get

$x^4+9x^2+36=\left(x^2+\dfrac92\right)^2+\dfrac{63}4$

So in the integral, you can substitute $y=x^2+\dfrac92$ to get

$\displaystyle\frac16\int_0^4x\sqrt{\left(x^2+\frac92\right)^2+\frac{63}4}\,\mathrm dx=\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy$

Next substitute $y=\dfrac{\sqrt{63}}2\tan z$ , so that the integral becomes

$\displaystyle\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy=\frac{21}{16}\int_{\arctan(3/\sqrt7)}^{\arctan(41/(3\sqrt7))}\sec^3z\,\mathrm dz$

This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):

$\displaystyle\int\sec^3z\,\mathrm dz=\frac12\sec z\tan z+\frac12\ln|\sec x+\tan x|+C$

So the arc length is

$\displaystyle\frac{21}{32}\left(\sec z\tan z+\ln|\sec x+\tan x|\right)\bigg|_{z=\arctan(3/\sqrt7)}^{z=\arctan(41/(3\sqrt7))}=\frac{21}{32}\ln\left(\frac{41+4\sqrt{109}}{21}\right)+\frac{41\sqrt{109}}{24}-\frac98$

4 0

4 years ago

Marlon asks a friend to think of a number from 5 to 11. What is the probability that Marlon’s friend will think of the number 9?

Strike441 [17]

Answer:

1/7

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Can you help me ?

kolezko [41]

It must have the [( )]

[(4/y) - z]

6 0

3 years ago

Nadia compares the weights, in grams, of some apples and oranges she find the median abs interquartile range of the weights.

choli [55]

Answer:

Apple

Orange

Step-by-step explanation:

Apple has the greater median with a median value of 150 which is more than that of orange at 130

The sample with the greater variability is the orange sample as the interquartile range value of 11 is greater Than 8 for Apple.

Yes it is possible, because of the greater variation on weight shown by the sample of oranges, then it could be possible.

3 0

3 years ago

If there consecutive terms of an AP are k+2 ,4k -6 and 3k-2 . What is the value of k

Natali5045456 [20]

<h3>Answer: k = 3</h3>

===========================================================

Explanation:

A = first term = k+2
B = second term = 4k-6
C = third term = 3k-2

To go from the first term to the second term, we add on some common difference d.

So,

B = A+d

B = (k+2)+d

4k-6 = k+2+d

4k-6-k-2 = d

d = 3k-8

---------------

Similarly, to go from the second term to the third term, we also add on d

C = B+d

C = (4k-6)+d

C = (4k-6)+(3k-8)

C = 7k-14

3k-2 = 7k-14

---------------

Let's solve for k

3k-2 = 7k-14

-2+14 = 7k-3k

12 = 4k

4k = 12

k = 12/4

k = 3 is the final answer

---------------

If k = 3, then we have these three terms:

A = k+2 = 3+2 = 5
B = 4k-6 = 4(3)-6 = 6
C = 3k-2 = 3(3)-2 = 7

The arithmetic progression (AP) is 5, 6, 7. The common difference is d = 1.

Note how d = 3k-8 = 3(3)-8 = 1

6 0

3 years ago