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3 years ago

9

Find the value of x and y

1 answer:

Amiraneli [1.4K]3 years ago

6 0

Answer:

y=110 because the angles correspond

x=55 because they are verticals angles and are congruent so you set them equal to each other

Explanation:

2x=110

x=55

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2 years ago

30 sweets are shared between 2 groups in the ratio 3:7. How many sweets are in each group?

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Answer:

See explanation below

Step-by-step explanation:

Given the ratio between two groups to be 3:7

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2 years ago

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1 year 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 $v(x)=\dfrac{y(x)}x$ :

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

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

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

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3 years ago

April shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high. The pathway of the

zaharov [31]

Answer:

The maximum height of the arrow is 125 feet.

Step-by-step explanation:

The pathway of the arrow can be represented by the equation,

$h = -16t^2 +80t + 25$ .....(1)

Where h is height in in feet and t is time in seconds.

It is required to find the maximum height of the arrow. For maximum height, $\dfrac{dh}{dt}=0$ .

So,

$\dfrac{d(-16t^2 +80t + 25)}{dt}=0\\\\-32t+80=0\\\\t=\dfrac{80}{32}\\\\t=2.5\ s$

Put t = 2.5 s in equation (1). So,

$h = -16(2.5)^2 +80(2.5) + 25\\\\h=125\ \text{feet}$

So, the maximum height of the arrow is 125 feet.

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2 years ago