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Anuta_ua [19.1K]
3 years ago
8

Natasha rolls a fair dice 66 times.

Mathematics
1 answer:
bonufazy [111]3 years ago
6 0
Im not sure but could be 60 times ?
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What is the power notation of 125
loris [4]

Answer:

5³

Step-by-step explanation:

125 = 5 × 5 × 5 = 5³

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3 years ago
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Put this equation into slope intercept -2×+3y=9
siniylev [52]

y = \frac{2}{3} x + 3

The equation of a line in slope-intercept form is

y = mx + c ( m is the slope and c the y-intercept )

rearrange - 2x + 3y = 9 into this form

add 2x to both sides

3y = 2x + 9 ( divide all terms by 3 )

y = \frac{2}{3} x + 3 ← in slope-intercept form


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3 years ago
Find the area of the figure
NikAS [45]

Answer:

14 square meters

Step-by-step explanation:

3*2=6

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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
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2 years ago
A translation is
Eduardwww [97]

Answer:

The answer is B..

Step-by-step explanation:

A translation is a transformation in which every point moves the same direction and distance.

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