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2 years ago

8

A circle with radius 9 has a sector with a central angle of 120°. What is the area of the sector. Provide step by step explanati

on please

1 answer:

inn [45]2 years ago

6 0

Okay, so you find the area of the full circle, 81pi and then 120 is 1/3 of the circle so you divide 81pi by 3 and get 27pi.

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Digiron [165]

Answer:

<h3>mean = 39+ 42+41+38+39+39+42/7</h3><h3>= 244 </h3>

6 0

3 years ago

What is a disadvantage of overpaying your taxes during the year?

Otrada [13]

The disadvantage of overpaying your taxes during the year is that you cannot use the money or draw interest on it until after you get your tax refund.

If you are overpaying your taxes, the government is holding that extra money for you. At the end of the tax year when you file your taxes, they will give you the extra money back that you overpaid.

However, you don't receive any interest on the extra money that was held.

5 0

2 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

I really don't understand why this question was deleted its not bad I don't know how to explain it

nalin [4]

Answer:

Step-by-step explanation:

Because it is a square root

8 0

2 years ago

Enter the value for x that makes the equation 3x – 5 = 7x - 21 true.

Nitella [24]

Answer:

4

Step-by-step explanation:

3x-5=7x-21

3x-7x=5-21

-4x=-16 ( divide by -4 both sides)

x= 4

8 0

2 years ago

Read 2 more answers