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3 years ago

PLEASE PLEASE PLEASE HELP ME

Mathematics

1 answer:

ziro4ka [17]3 years ago

7 0

Answer:

276 .........................................

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3.x4 - 2x2 + 15.x2 Which product Is the factored form of the polynomial

Dmitry_Shevchenko [17]

Answer:

3*4=12

12-2=10

10*2=20

20+15=35

35*2=70

Step-by-step explanation:

4 0

3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

Write the equation in spherical coordinates. (a) x2 y2 z2 = 64

vekshin1

The equation in spherical coordinates will be a constant, as we are describing a spherical shell.

r(φ, θ) = 8 units.

<h3>How to rewrite the equation in spherical coordinates?</h3>

The equation:

x^2 + y^2 + z^2 = R^2

Defines a sphere of radius R.

Then the equation:

x^2 + y^2 + z^2 = 64

Defines a sphere of radius √64 = 8.

Then we will have that the radius is a constant for any given angle, then we can write r, the radius, as a constant function of θ and φ, the equation will be:

r(φ, θ) = 8 units.

If you want to learn more about spheres, you can read:

brainly.com/question/10171109

8 0

2 years ago

ABCD is rectangle. Find the length of each diagonal AC=c/3 BD=4-c

Vikki [24]

The lenght of each diagonals is 1.

3 0

3 years ago

Read 2 more answers

-4(9n + 9) = -12(4n - 6).

WITCHER [35]

Answer:

n = 9

Step-by-step explanation:

Step 1: Write equation

-4(9n + 9) = -12(4n - 6)

Step 2: Solve for n

Distribute: -36n - 36 = -48n + 72

Add 48n to both sides: 12n - 36 = 72

Add 36 to both sides: 12n = 108

Divide both sides by 12: n = 9

Step 3: Check

Plug in n to verify if it's a solution.

-4(9(9) + 9) = -12(4(9) - 6)

-4(81 + 9) = -12(36 - 6)

-4(90) = -12(30)

-360 = -360

7 0

3 years ago