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3 years ago

4xy- 72 = 3w solve for y

Mathematics

1 answer:

Mashcka [7]3 years ago

6 0

Answer:

y= $\frac{3w}{4x}$ + $\frac{18}{x}$

Step-by-step explanation:

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Find the mean of this data: 49,49,56,62,60,50,59 if necessary, round to the nearest tenth

aleksandrvk [35]

(49+49+56+62+60+50+59) / 7 = 55

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3 years ago

WILL MARK BRAINLIEST What is the value of j?

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3 years ago

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A computer is programmed to scan the digits of the counting numbers. For example, if it scans 1 2 3 4 5 6 7 8 9 10 11 12 13 then

Charra [1.4K]

I believed it is "401"

1-9 = 9 digits

10-99 = 90 which is 180

100-200 = 100 which is 300 digits

9+180= 189

1392-189= 1203

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401 is the 1392 number scanned

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3 years ago

Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect

Tasya [4]

$4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1$

defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

$\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}$
$\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}$

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

and so the volume would be

$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$

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3 years ago

Megan bikes west to get from her apartment to school after school she bikes 8 miles north to her friends house how far is megans

snow_tiger [21]

The complete question: Megan bikes 6 miles west to get from her apartment to school. After school, she bikes 8 miles north to her friend's house. How far is Megan's apartment from her friend's house, measured in a straight line?

So,Megan's apartment, the school, and Megan's friend's house form a right triangle with legs 6 miles and 8 miles as you can see in the attached picture; the hypotenuse of the right triangle is the straight distance is from Megan's apartment to her friend's house. To find that distance we are going to use the Pythagorean theorem:
 $h= \sqrt{(6mi)^2+(8mi)^2}$
$h= \sqrt{36mi^2+64mi^2}$
$h= \sqrt{100mi^2}$
$h=10mi$

We can conclude that distance between Megan's apartment and her friend's house is 10 miles.

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3 years ago