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Jobisdone [24]
3 years ago
8

Please help with the question attached, will mark brainliest!!!!!!!!!!!

Mathematics
1 answer:
UkoKoshka [18]3 years ago
7 0

12; 4

66; 33.4

32.6

respectively in the blanks

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The backboard of the basketball hoop forms a right triangle with the supporting rods, as shown. Use the
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Answer:

9.14 inches.

Step-by-step explanation:

From the question,

Assuming the diagram attached, is similar to diagram required to support the question,

From the diagram,

Applying pythagoras theorem

a² = b²+c²...................... Equation 1

Where a = 13.4 in, b = x in, c = 9.8 in.

Substitute these values into equation 1

13.4² = x²+9.8²

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The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.
Taya2010 [7]
Given a solution y_1(x)=\ln x, we can attempt to find a solution of the form y_2(x)=v(x)y_1(x). We have derivatives

y_2=v\ln x
{y_2}'=v'\ln x+\dfrac vx
{y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}

Substituting into the ODE, we get

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w'x\ln x+(2+\ln x)w=0

Multiplying both sides by \ln x, we have

w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0

and noting that

\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x

we can write the ODE as

\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0

Integrating both sides with respect to x, we get

wx(\ln x)^2=C_1
w=\dfrac{C_1}{x(\ln x)^2}

Now solve for v:

v'=\dfrac{C_1}{x(\ln x)^2}
v=-\dfrac{C_1}{\ln x}+C_2

So you have

y_2=v\ln x=-C_1+C_2\ln x

and given that y_1=\ln x, the second term in y_2 is already taken into account in the solution set, which means that y_2=1, i.e. any constant solution is in the solution set.
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