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2 years ago

Who play's Little big snake "https://littlebigsnake.com/"

Mathematics

1 answer:

daser333 [38]2 years ago

5 0

Answer:

I have never heard of it sorry

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C=48y+0.99n solve for y

Ne4ueva [31]

Answer:

y = (C -0.99n)/48

Step-by-step explanation:

Undo what is done to y, in the reverse order. We have y multiplied by 48 then added to 0.99n. So, we must add the opposite of 0.99n and multiply that result by the reciprocal of 48.

C = 48y + 0.99n . . . . . . given

C - 0.99n = 48y . . . . . . add -0.99n

(C -0.99n)/48 = y . . . . . multiply by 1/48

4 0

3 years ago

Find the surface area of a cube with the following side length of 5<br><br> sa=

laiz [17]

Answer:

150 units^2

Step-by-step explanation:

A=6a2=6·52=150

5 0

3 years ago

Read 2 more answers

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

2 years ago

Which could a translation result in? Select all that apply. a figure getting smaller a figure moving up a figure flipping upside

makkiz [27]

Answer:

A figure moving up, a figure moving left, and a figure moving down.

Step-by-step explanation:

Translations are just moving things around in any direction.

8 0

2 years ago

Read 2 more answers

A man invests $2000 in an account that pays 6.5% interest per year, compounded quarterly.

skelet666 [1.2K]

Answer:

The correct answer is a. $2275.28; b. 17.04 years

Step-by-step explanation:

Principal to be invested is $2000

Interest rate (r) per year is 6.5 quarterly.

Interest is calculated compoundly.

a. Time (t) for the investment is given to be 2 years.

Amount after two years is = Principal × $( 1+ \frac{r}{100n}) ^ {nt}$ where the value of n is 4.

⇒ A = 2000 × $( 1+ \frac{6.5}{400}) ^ {8}$

⇒ A = $2275.28.

b. Now the value of A is given to be triple the principal = $ (3 × 2000).

Therefore we need to find the value of t.

⇒ 3 × 2000 = 2000 × $( 1+ \frac{6.5}{400}) ^ {4t}$

⇒ ㏑ 3 = 4t × ㏑ ( 1.01625)

⇒ t = 17.04

Therefore it would take 17.04 years for the principal to triple.

5 0

2 years ago