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jasenka [17]
3 years ago
8

A round pen is 50ft in diameter, what is the circumference of the round pen?

Mathematics
2 answers:
Pepsi [2]3 years ago
6 0
C=πd
d=50 ft
C=50π
Or
C=50×3.14
C=157 ft. Hope it help!
gizmo_the_mogwai [7]3 years ago
3 0
The answer would be 157.07, with a radius of 25 


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5x$12=$60 (spent on scarves)
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Pls help asap!!! just need answers not work
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Answer:

b= 5\sqrt{3} c = 10,   x=7 hypotenuse = 7\sqrt{2}

Step-by-step explanation:

A 30-60-90 triangle has the ratios of x-x\sqrt{3}-2x so if x=5, then 5-5\sqrt{3} - 10

for a 45-45-90, if one leg is 7, so is the other x\sqrt{2} so hypotenuse is 7\sqrt{2}

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Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable
bekas [8.4K]

Answer:

\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

u=\frac{y}{x}

Multiply both sides by x:

ux=y

Second step: Differentiate the substitution:

u'x+u=y'

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}

u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of x^2 so I'm going to divide top and bottom by x^2:

u'x+u=\frac{u+u^2}{3+u^2}

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

u'x=\frac{u+u^2}{3+u^2}-u

Find a common denominator: Multiply second term on right hand side by \frac{3+u^2}{3+u^2}:

u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}

Combine fractions while also distributing u to terms in ( ):

u'x=\frac{u+u^2-3u-u^3}{3+u^2}

u'x=\frac{-u^3+u^2-2u}{3+u^2}

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

u' \frac{3+u^2}{-u^3+u^2-2u}x=1

Divide both sides by x:

\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}

Reorder the top a little of left hand side using the commutative property for addition:

\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0
3 years ago
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