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2 years ago

Please help with this question!! Will mark brainliest!! Thank you:)

Mathematics

1 answer:

madam [21]2 years ago

7 0

Answer:

I’m not good at algebra but I think that qx = -4 and qy = -7? I tried I’m not good at algebra.

Step-by-step explanation:

qy= -7 plus px= 3 equals -4

im not good at algebra

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Sidney skated for 1 1/10 hours last month and 8 7/10 hours this month. How many

olganol [36]

Answer:

the number of hours does sidney skate in all is 9.8 hours

Step-by-step explanation:

The computation of the number of hours does sidney skate in all is as follows:

= Last month + this month

$= 1 \frac{1}{10} + 8 \frac{7}{10}\\\\= \frac{11}{10} + \frac{87}{10}\\\\= \frac{98}{10}$

= 9.8 hours

Hence, the number of hours does sidney skate in all is 9.8 hours

3 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

2 years ago

What are the solutions to the system of equations graphed

sweet-ann [11.9K]

Answer:

The point of intersection between the 2 graphs.

5 0

3 years ago

12598 round to the nearest thousand

MakcuM [25]

12598 rounded to the nearest thousands is 1260

5 0

3 years ago

Read 2 more answers

1. What is the slope intercept form for -3, and (4,2) SIMPLIFIED??

Ostrovityanka [42]

Answer:

Step-by-step explanation:

1. y - 2 = -3(x - 4)

y - 2 = -3x + 12

y = -3x + 14

2. y - 6 = -1/2(x + 5)

y - 6 = -1/2x - 5/2

y - 12/2= -1/2x - 5/2

y = -1/2x + 7/2

4 0

2 years ago

Please help with this question!! Will mark brainliest!! Thank you:)​

Please help with this question!! Will mark brainliest!! Thank you:)