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Goryan [66]
2 years ago
12

Find the value of x.

Mathematics
1 answer:
allsm [11]2 years ago
8 0

Answer:

x=51

Step-by-step explanation:

Formula:  35+16=x

Step 1: Simplify both sides of the equation.

35+16=x

(35+16)=x(Combine Like Terms)

51=x

51=x

Step 2: Flip the equation.

x=51

Hope this helps.

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Find the 22nd term of the following sequence:<br> 5, 8, 11, ...<br> 63 <br> 71<br> 14<br> 68
Wittaler [7]

68 is the 22nd term of the following sequence.

<u>Step-by-step explanation:</u>

  • The given sequence is with the same common difference between the two consecutive number in the series thus it is said to be the  Arithmetic progression ( AP).
  • For finding the nth term in the AP we have a formula tn = a + (n-1) × d
  • Here a is the first term , n is the number of the term to be founded and d is the common difference between the two consecutive number in the series.
  • Thus here tn = 5 + ( 22 - 1 ) × 3.
  • On subtracting we get tn = 5 + (21 ) × 3
  • On multiplying we get tn = 5 + 63
  • After adding we get tn = 68. It is the 22nd term in the given series.
5 0
3 years ago
Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x
djverab [1.8K]

I'll assume the ODE is

y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}

Solve the homogeneous ODE,

y'' - 3y' + 2y = 0

The characteristic equation

r^2 - 3r + 2 = (r - 1) (r - 2) = 0

has roots at r=1 and r=2. Then the characteristic solution is

y = C_1 e^x + C_2 e^{2x}

For nonhomogeneous ODE (1),

y'' - 3y' + 2y = e^x

consider the ansatz particular solution

y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x

Substituting this into (1) gives

a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1

For the nonhomogeneous ODE (2),

y'' - 3y' + 2y = e^{2x}

take the ansatz

y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}

Substitute (2) into the ODE to get

b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1

Lastly, for the nonhomogeneous ODE (3)

y'' - 3y' + 2y = e^{-x}

take the ansatz

y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}

and solve for c.

ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16

Then the general solution to the ODE is

\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}

6 0
1 year ago
If every minute i geh 4.5 million yen, how much would i get in 24 hours.
hammer [34]

Step-by-step explanation:

hear is your answer in attachment

3 0
2 years ago
Again.. your girl needs help again ..
Maru [420]

<u>Answer:</u>  C. 3x^{2} + 3

<u>Explanation:</u>  you basically combine f(x) and g(x).

2x^{2} + 5 + x^{2} -2      combine your terms

3x^{2} + 2    your final answer


hope this helps!❤ from peachimin


6 0
3 years ago
Graph the line with slope −1/2 and y-intercept −5.
Allushta [10]

First, we can write the equation of the line using the information provided:

\begin{gathered} y=mx+b \\ where: \\ m=slope=-\frac{1}{2} \\ b=y-intercept=-5 \\ so: \\ y=-\frac{1}{2}x-5 \end{gathered}

Now, we can create a table:

Finally, we can graph the line:

5 0
1 year ago
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