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3 years ago

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Mathematics

1 answer:

kvv77 [185]3 years ago

3 0

Answer:

Step-by-step explanation:

yellow was picked more times then green so the percent would be more then 50.

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4x-2x what is the answer to this?

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2x is the answer to that question it's simple math just

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2 years ago

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Use the method of completing the square to transform the quadratic equation into the equation form (x + p)^2 =q using the equati

fenix001 [56]

Answer:

Step-by-step explanation:

6+12x+3x²=0

3x²+12x=-6

x²+4x=-2

x²+4x+4=-2+4

(x+2)²=2

p=2

q=2

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2 years ago

Simone claims the initial value and y-intercept are the same thing on a graph. Is she correct? If you know that a line has a slo

alexandr1967 [171]

Answer:

Yes, she is right.

in y = m*x + b

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If you know that a line has a slope of m (was not given) and a y-intercept of 7, what is the equation for that line in slope-intercept form?

y = m*x + 7

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3 years ago

Solve the differential equation <img src="https://tex.z-dn.net/?f=y%20%7B%7D%5E%7B%285%29%7D%20-4y%20%7B%7D%5E%7B%284%29%7D%2

Romashka-Z-Leto [24]

The given differential equation has characteristic equation

$r^5 - 4r^4 + 4r^3 - r^2 + 4r - 4 = 0$

Solve for the roots $r$ .

$r^3 (r^2 - 4r + 4) - (r^2 - 4r + 4) = 0$

$(r^3 - 1) (r^2 - 4r + 4) = 0$

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

$r^3 - 1 = 0 \text{ or } (r-2)^2=0$

The first case has the three cubic roots of 1 as its roots,

$r^3 = 1 = 1e^{i0} \implies r = 1^{1/3} e^{i(0+2\pi k)/3} \text{ for } k\in\{0,1,2\} \\\\ \implies r = 1e^{i0} = 1 \text{ or } r = 1e^{i2\pi/3} = -\dfrac{1+i\sqrt3}2 \text{ or } r = 1e^{i4\pi/3} = -\dfrac{1-i\sqrt3}2$

while the other case has a repeated root of

$(r-2)^2 = 0 \implies r = 2$

Hence the characteristic solution to the ODE is

$y_c = C_1 e^x + C_2 e^{-(1+i\sqrt3)/2\,x} + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}$

Using Euler's identity

$e^{ix} = \cos(x) + i \sin(x)$

we can reduce the complex exponential terms to

$e^{-(1\pm i\sqrt3)/2\,x} = e^{-x/2} \left(\cos\left(\dfrac{\sqrt3}2x\right) \pm i \sin\left(\dfrac{\sqrt3}2x\right)\right)$

and thus simplify $y_c$ to

$y_c = C_1 e^x + C_2 e^{-x/2} \cos\left(\dfrac{\sqrt3}2x\right) + C_3 e^{-x/2} \sin\left(\dfrac{\sqrt3}2x\right) \\ ~~~~~~~~ + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}$

For the non-homogeneous ODE, consider the constant particular solution

$y_p = A$

whose derivatives all vanish. Substituting this into the ODE gives

$-4A = 69 \implies A = -\dfrac{69}4$

and so the general solution to the ODE is

$y = -\dfrac{69}4 + C_1 e^x + C_2 e^{-x/2} \cos\left(\dfrac{\sqrt3}2x\right) + C_3 e^{-x/2} \sin\left(\dfrac{\sqrt3}2x\right) \\ ~~~~~~~~ + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}$

3 0

1 year ago

Y= 12x+2 2y = x+4 and the step by step explanation

Lena [83]

Answer:

$x=0, y=2$

Step-by-step explanation:

Below! Credits to WolframAlpha!

5 0

3 years ago