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Mrac [35]
3 years ago
8

NEED THE EQUATION

Mathematics
1 answer:
katrin [286]3 years ago
7 0

Answer:

d

Step-by-step explanation:

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How do i solve this using imaginary numbers? PLEASE HELP
Tasya [4]

Answer:

14 -2i

Step-by-step explanation:

9 + sqrt(-4) - ( -5 + sqrt(-16))

We know the sqrt of a negative number is i * sqrt(number)

9 + isqrt(4) - ( -5 + isqrt(16))

9 + 2i - (-5 + 4i)

Distribute the negative sign

9+2i +5 -4i

Combine like terms

14 -2i

8 0
3 years ago
What values of b satisfy 3(26 + 3)² = 36?
MakcuM [25]

The numeric value of the b is 0.23 which satisfy the given equation.

According to the statement

We have to find that the value of the b.

So, For this purpose, we know that the

The numeric value refers to the worth of each digit depending on where it lies in the number.

From the given information:

The equation is a

3(2b + 3)² = 36

then to find the value of b rearrange the terms then

(2b + 3)² = 12

(2b + 3) =  \sqrt{12}

(2b) =   \sqrt{12}  -3

then

b  = \sqrt{12}  -3/ 2

b = 3.46 - 3/2

b = 0.46/2

b = 0.23.

So, The numeric value of the b is 0.23 which satisfy the given equation.

Learn more about numeric value here

brainly.com/question/24703884

#SPJ9

7 0
1 year ago
Anyone know the answer?
Anuta_ua [19.1K]
21x - 19 = 3x + 17   |add 19 to both sides
21x = 3x + 36    |substract 3x from both sides
18x = 36     |divide both sides by 18
x = 2

7 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
How many solutions to this problem
zlopas [31]
None
:::::::::::::::::::::
4 0
3 years ago
Read 2 more answers
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