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2 years ago

What will happen if Pinocchio says "my nose will grow now" ??

Mathematics

1 answer:

Anna [14]2 years ago

5 0

Maybe he lies under his breath right after saying that, then his nose will grow.

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Mr. Sonny's science class is calculating the average number of blinks per minute. Jan blinks 57 times in 3 minutes. What is her

Vlada [557]

b. Every 24 minutes

Step-by-step explanation:

A neon sign blinks every 6 minutes, and another sign blinks every 8 minutes.

The first step is to:

Find and list multiples of each time each neon sign blinks.

This means, we find the Multiples of 6 and 8

Multiples of 6:

6, 12, 18, 24, 30, 36

Multiples of 8:

8, 16, 24, 32, 40

The first and common multiple of 6 and 8.

Therefore, the Least common multiple [LCM(6, 8) ] = 24

Hence, they often blink at the same time every 24 minutes.

Therefore, option b.Every 24 minutes is correct.

please mark me brainliest!

3 0

2 years ago

Read 2 more answers

The soccer team and the American football team are sharing the field for practice today. The soccer team has practice every 2 da

Maru [420]

I believe the answer would be 4 days.

4 0

3 years ago

Read 2 more answers

3) What is an equation for the line that passes through the coordinates (-1.2) and (7,6) ?

OLEGan [10]

Answer:

$y=\frac{1}{2} x+\frac{5}{2}$

Step-by-step explanation:

First find the slope using the slope using the given points. Remember that the slope is the change in y over the change in x.

$\frac{6-2}{7- (-1)}=\frac{4}{8}=\frac{1}{2}$

So now we have the equation $y=\frac{1}{2} x+b$ , and we need to find out what b is. We can do this by pointing a point (either one, but I'll use -1,2) into the equation

$2=\frac{1}{2}(-1)+b$

Rearrange the equation so it equals b

$\frac{5}{2}=b$

Put it together and that's the final equation!

$y=\frac{1}{2} x+\frac{5}{2}$

7 0

2 years ago

The apple pie baked from 8:25 P.M. until 8:55 P.M. How much time did it take the pie to bake?

maw [93]

Answer:

30 minutes.

Step-by-step explanation:

8:55 - 8:25 is 30 minutes.

If you add 30 to 8:25, you would get 8:55.

Therefore, it's 30 minutes.

8 0

3 years ago

Read 2 more answers

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

2 years ago