How to write 54 hundredths

Y is directly proportional to x.

Anna35 [415]

Answer: Our required equation would be

Step-by-step explanation:

Since we have given that

y=30, when x = 6.

As there is direct proportion between x and y.

so, it becomes,

So, the equation connecting y and x would be

When we put the value of k, we get that

Hence, our required equation would be y=5x

5 0

1 year ago

Why have I been blocked?

Ivanshal [37]

I’m not sure blocked from what

8 0

2 years ago

Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

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 $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

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

Finally, the solution is given by:

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

5 0

3 years ago

Find an equation for the line below.

tekilochka [14]

Answer:

y=3/4x-2.5

Step-by-step explanation:

This is a straight line, so it would follow the equation y=mx+b. You can find the b value by looking at the y-intercept of the equation (where x=0), which in this case would be -2.5. Then, you can find the m value by comparing the two given points' locations. Slope is rise over run, so first you can find the rise. The y value moves from y=-4 to y=2, which is a value of 6 units upward. The points move from x=-2 to x=6, which is a distance of 8 units. Finally, put it into the rise over run pattern, giving you 6/8, which simplified is 3/4. This is your m value. Finally, plug in your two variable values into y=mx+b and you'll get y=3/4x-2.5.

7 0

3 years ago

HELPPPPPP ASAPPPPPPP!!!!!!!!!!!!!!!!!!!!

Burka [1]

Answer:

10(x+y)

Step-by-step explanation:

the sum is addition

4 0

2 years ago