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3 years ago

Please help me with these two problems thank you!!

Mathematics

1 answer:

andriy [413]3 years ago

8 0

Answer:

1st question: y=1/3x-1

2nd question: y=-2/3x+9

Step-by-step explanation:

Y=mx+b form is what you are using

when a line in parallel to a line they have the same slop wich is m

when a line in perpendicular they have the opposite reciprocal

Examples: 2/5 opposite reciprocal would be -5/2

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What is 3 ÷ 6 equal as a fraction

leva [86]

Step-by-step explanation:

1/2 is the answer

hope it helps

thank you

7 0

2 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

Tammy bought a nightstand that was marked down 60% from an original price of $50. If she paid 10% sales tax, what was the total

AURORKA [14]

Answer: the total cost of the nightstand is $22

Step-by-step explanation:

The original price of the night stand is $50. If the original price of the night stand was marked down by 60%, it means that the amount of money by which it was reduced would be

60/100 × 50 = 0.6 × 50 = 30

The new selling price would be

50 - 30 = $20

If she paid 10% sales tax,the amount of tax paid would be

10/100 × 20 = 0.1 × 20 = 2

Therefore, the total cost of the nightstand would be

20 + 2 = $22

7 0

3 years ago

1 2 3 4 .... 1002 what is the sum of the sequence

Reika [66]

<span>As far as i know it is related to Gauss.
Write the sequences forward and backward first.

1 +2 +3 +.....+1002
1002+1001+1000+.....+1
--------------------------------------... Adding them
1003+1003+......(1002 times)
=1002x1003
But this contains the series twice.
So, the sum is = 1002x1003/2=501x1003=502503. answer</span>

3 0

2 years ago

-8 3/8+6 1/4 written in simplest form

natka813 [3]

Step-by-step explanation:

-8 3/8 + 6 1/4 = -2 1/8 or you can also write -17/8

3 0

2 years ago

Please help me with these two problems thank you!!​

Please help me with these two problems thank you!!