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Misha Larkins [42]
2 years ago
5

Help me! Help me! HEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLPPPPPPPPPPP MEEEEEEEEEEEEEEEEEEE

Mathematics
2 answers:
oksian1 [2.3K]2 years ago
5 0

Answer:

uhhhhhhhhhhh...............

bija089 [108]2 years ago
4 0

Answer:

no.

Step-by-step explanation:

amogus sus bobux man

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What is 100.2 percent as a fraction and decimal
notka56 [123]

Answer:

1.002 Decimal and 1 2/100 fraction

Step-by-step explanation:

Move the decimal 2 places to the right and make 100.2 1.002

think 100.2 over 100. we get 100 2/100 also know as 1whole and 2/100

7 0
3 years ago
Read 2 more answers
Solve for y.<br> y y<br> —— = ——<br> P-9 P+9
Veronika [31]

9514 1404 393

Answer:

  y = 0

Step-by-step explanation:

Subtracting the right side gives ...

  \dfrac{y}{P-9}-\dfrac{y}{P+9}=0\\\\y\left(\dfrac{(P+9)-(P-9)}{P^2-81}\right)=0\\\\y\left(\dfrac{18}{P^2-81}\right)=0\\\\\boxed{y=0}

The factor multiplying y cannot be zero, so y must be zero.

5 0
3 years ago
An ice cream shop offers 3 ice cream flavors 3 types of syrup and 2 toppings for sundaes
babymother [125]

Answer: 18

Step-by-step explanation:

3 0
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
Can someone help me with this question?
Ilia_Sergeevich [38]

Step-by-step explanation:

hope it helps, I have chegg and they seem to have more answers and a lot of topics. you can scan your books bar code and get end of chapter questions

3 0
2 years ago
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