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steposvetlana [31]

3 years ago

13

A footbridge is in the shape of an arc of a circle. The bridge is 7 ft tall and 29 ft wide. What is the radius of the circle tha

t contains the bridge? Round your answer to the nearest tenth.

1 answer:

Mashcka [7]3 years ago

5 0

Answer:

18.5 ft (to the nearest tenth)

Step-by-step explanation:

The diameter of the circle passes through the center of the circle and divides the width of the circle into two equal parts.

We complete the circle and apply the <u>Law of Intersecting Chords.</u>

In the diagram, by this law:

AB X BC =DB X BE

7*x=14.5X14.5

x=14.5X14.5÷7

x=30.04

Therefore:

Length of the Diameter= 7+30.04 ft=37.04 ft

Radius =37.04/2= 18.5 ft

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Find all the solutions for the equation:

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

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

Coach Martinez will order 2 pairs of shorts and 3 shirts for each player. There are 12 members on the team. If each pair of shor

artcher [175]

Answer:

12(3y+2x)

Step-by-step explanation:

In this question, we are asked to give an expression that represents the total cost of shorts for the players.

He is ordering 2 pairs of short and 3 shirts. Hence, per player, he would be spending the following:

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Per player he is spending a total of 3y + 2x

Now, there are people to cater for. The total amount to spend will now be 12(3y + 2z )

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Japan has placed tariffs on foreign goods which allows consumers to pay lower prices because goods are produced domestically.

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Y=3x-5 y=-6x+4 system of equations

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