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3 years ago

Tim is 3 times as old as chris and is also 14 years older than chris. how old is tim?

Mathematics

1 answer:

Luda [366]3 years ago

7 0

Let y represent Tim's age and let x represent Chris's age.

y=3x

y=x+14

Set the two equations equal to each other:

3x=x+14

Subtract the x from the one side to get it on the other side.

2x=14

Divide by 2 on both sides

14/2=7

x=7

Plug it back in and see if it works:

y=3(7)=21

Other equation:

7+14=21

x=7

y=21

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2 years ago

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Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

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$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

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3 years ago

Find the cost of leveling a badminton court 132m long and 58 m broad at the rate of Rs. 80 per sq. m.

serg [7]

Answer:

Rs. 612,480

Step-by-step explanation:

Hi there,

Here the question asks us to find the cost of leveling a badminton court at the rate of Rs.80 per sq..

So by the word leveling we know that we have to find the area of the court in order to proceed further.

(I am assuming the court to be rectangular in shape).

Area of a rectangle = Length * Breadth

*Length = 132 meters

*Breadth = 58 meters

==> 132*58= 7656 m^2

So now that we got the area of the badminton court lets find the cost of leveling it ==>

Cost of leveling per meter^2 = Rs. 80

Area = 7656 m^2

==> 7656 * 80 = 612,480

So the cost would be Rs. 612,480

If you found this answer helpful please mark me as brainliest.

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