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3 years ago

HELP ME I'M STUCK!!!! PLEASE AND THANK YOU!!!!

Mathematics

1 answer:

Darya [45]3 years ago

3 0

Answer: 1/8

Step-by-step explanation: The sample space are all of the different outcomes an event can lead to. The H and T combinations at the bottom are all of the different outcomes. I think the probability of getting 3 tails for the coins (or TTT) is 1/8. This is because there are 24 total outcomes (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT). There is 3/24 TTTs (also, 1/8 if you count the groups/simplify). Therefore, it is 1/8. Hope this helps!

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I really help with this ill mark u brainliest pls

erma4kov [3.2K]

Answer:

$m\angle XYW = (155 - 20x)\degree$

Step-by-step explanation:

$m\angle XYZ= m\angle XYW +m\angle WYZ\\\\150\degree = m\angle XYW +(20x-5)\degree \\\\150\degree - (20x-5)\degree = m\angle XYW \\\\(150 - 20x+5)\degree = m\angle XYW \\\\(155 - 20x)\degree = m\angle XYW \\\\m\angle XYW = (155 - 20x)\degree \\\\$

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3 years ago

What is the slope of the line that modles the table below?

Shtirlitz [24]

Answer:

3/4

Step-by-step explanation:

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3 years ago

18.) Which expression is equivalent to -6x + 7.5?

nydimaria [60]

Answer:

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3 years ago

Twice the sum of a number and 8 is the same aa the difference of 3 and the number. Find the equation that could be used to find

Nezavi [6.7K]

Answer:

f your number is x, then the difference between that number and 8 would be x-8. Twice that is 2(x-8)

Three times the sum of the number and 3 would be 3(x+3). So, you get the equation:

2(x-8)=3(x+3)

2x-16=3x+9

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3 years ago

Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).

I am Lyosha [343]

Take the homogeneous part and find the roots to the characteristic equation:

$y''+4y=0\implies r^2+4=0\implies r=\pm2i$

This means the characteristic solution is $y_c=C_1\cos2x+C_2\sin2x$ .

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form $y_p=ax\cos2x+bx\sin2x$ . Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With $y_1=\cos2x$ and $y_2=\sin2x$ , you're looking for a particular solution of the form $y_p=u_1y_1+u_2y_2$ . The functions $u_i$ satisfy

$u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$

where $W(y_1,y_2)$ is the Wronskian determinant of the two characteristic solutions.

$W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2$

So you have

$u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx$
$u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x$

$u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx$
$u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x$

So you end up with a solution

$u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

but since $\cos2x$ is already accounted for in the characteristic solution, the particular solution is then

$y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x$

so that the general solution is

$y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

7 0

3 years ago