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Reika [66]
2 years ago
6

Pls help me 4x-2y=14 y=1/2x-1

Mathematics
1 answer:
kati45 [8]2 years ago
5 0
Judging by the problem you want me to use substitute method
Since why equals 1/2x-1 you
Do 2(1/2x-1)= x-2
You multiple by 2 cause there’s two on the top for y. So the problem is now 4x-x-2
Combine like terms 3x+2=14
Subtract two on both sides
3x=12
X=4
Go back to the bottom since u know x now do y= 4.1/2-1 solve to get one
Y=1 so your answer is (4,1)

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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
(10 points for the answer)<br><br> y=
Nat2105 [25]
Answer: y=1x+4
Solving Steps:
5 0
3 years ago
Which experiment describes a binomial experiment?
marshall27 [118]
Answer: the last option.
A binomial experiment is an experiment with only two possible outcomes.

In the first option, there is three possible outcomes (i.e. A, B, C)
In the second option, there is four poissible outcomes (i.e. ace, king, queen, jack)
The third option has six possible outcomes (i.e. six faces of a die)
In the last option, there are two possible outcomes (i.e. '2 number cubes')
6 0
3 years ago
Cube-shaped blocks are packed into a cube-shaped storage container. The edge of the storage container is 2 1/2 feet. The edge le
GarryVolchara [31]

Answer:

Volume of one cube shaped block is, 0.125 cubic feet

Step-by-step explanation:

Volume of a cube(V) is given by:

V = a^3                 .....[1]

where a is the edge length of the cube.

As per the statement:

Cube-shaped blocks are packed into a cube-shaped storage container. The edge of the storage container is 2 1/2 feet.

⇒\text{Edge length of storage} = 2\frac{1}{2} = 2.5 ft

It is also given that:

The edge length of each block is 1/5 the edge length of the storage container

⇒\text{Edge length of each block (a)} = \frac{1}{5} \cdot 2.5 = 0.5 ft

Substitute this in [1] we have;

V = (0.5)^3 = 0.125 ft^3

Therefore,  volume, in cubic feet, of one cube-shaped block is, 0.125

4 0
3 years ago
Let f(x) be an exponential functionWhich transformations to the graph of f(x) would describe e - f(x - 3) ?
gayaneshka [121]

Look at the graph below carefully

Observe the results of shifting ={2}^{x}f(x)=2​x

​​  vertically:

The domain, (−∞,∞) remains unchanged.

When the function is shifted up 3 units to ={2}^{x}+3g(x)=2​x +3:

The y-intercept shifts up 3 units to (0,4).

The asymptote shifts up 3 units to y=3y=3.

The range becomes (3,∞).

When the function is shifted down 3 units to ={2}^{x}-3h(x)=2 ​x​​ −3:

The y-intercept shifts down 3 units to (0,−2).

The asymptote also shifts down 3 units to y=-3y=−3.

The range becomes (−3,∞).

3 0
2 years ago
Read 2 more answers
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