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Svetradugi [14.3K]
3 years ago
13

I need to know the answer quick

Mathematics
1 answer:
Paha777 [63]3 years ago
3 0

Answer:

v = 13

Step-by-step explanation:

Truly hope I helped fast enough:).

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Please answer question 17. Explain the steps. (A bit urgent though)
gregori [183]
Angle a is 80, angle d is 100, angle c is 100

6 0
3 years ago
What is the solution of this system of linear equations?<br><br> 3y = x + 6<br><br> y – x = 3
Naya [18.7K]
<span>3y = x + 6
y – x = 3 ----------->  x=y - 3 (substitute in 1st equation)

3y = y - 3 +6
3y - y = 3
2y = 3
y = 2/3

x= y - 3 = 2/3 -3 = 2/3 - 9/3 =  - 7/3

</span>x= - 7/3, y = 2/3<span>


</span>
4 0
3 years ago
Read 2 more answers
Find the distance of point U(2, 3) from point V(5, 7).​
aalyn [17]

Answer:

5

Step-by-step explanation:

You can draw a point at (5,3) and call it point A.

Connect point A with V and U to get VA = 4 and UA = 3.

You get a right triangle and use pythagoream theroem. The answer is the root of 4^2 + 3^2 which equals to 5.

The answer is 5

5 0
3 years ago
Emily buys a dress for ?48 and adds 25% onto the price before putting it into her shop window what price does he price tag say i
Maurinko [17]

Answer:

$60

Step-by-step explanation:

$48 x 25% = $12

$48+$12 =$60

6 0
3 years ago
Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).
Firdavs [7]

Answer:

A) Particular solution:

2x+\frac{1}{2}e^{-x}-\frac{8}{5}

B) Homogeneous solution:

y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))

C) The most general solution is

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

Step-by-step explanation:

Given non homogeneous ODE is

y''+4y'+5y=10x+e^{-x}---(1)

To find homogeneous solution:

D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)

To find particular solution:

y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\

Substituting y_{p},y'_{p},y''_{p} in (1)

y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\

Equating the coefficients

5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\

The general solution is

y=y_{h}+y_{p}

from (2) ad (3)

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

6 0
3 years ago
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