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3 years ago

Which of these expressions shows unlike terms?

Mathematics

1 answer:

emmainna [20.7K]3 years ago

6 0

Answer:

2x+2y

Step-by-step explanation:

x and y are unlike terms, and the only unlike terms in this list. you cannot add them together – therefore, 2x+2y is the answer you seek.

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How many solutions does the equation have? show your work.

Inessa05 [86]

All real numbers

7w-(2+w) = 2(3w-1)

Expand

7w-(2+w)= 6w-2

7w-2-w=6w-2

Group like terms

6w-2=6w-2

Add 2 to both sides

6w=6w

subtract 6w from both sides

0=0

True for all w

6 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

Based on the family the graph below belongs to, which equation could represent the graph?

Ostrovityanka [42]

Answer:

Equation 1 is the equation which represents the graph.

Step-by-step explanation:

6 0

2 years ago

Worth 20 points, please help and give good answers than you(will mark brainlest)

nalin [4]

Answer:

Step-by-step explanation:

all you need is to solve the first term

2x² - (-3x²) = 5x² ,

b is the only answer with 5x² in it

4 0

2 years ago

Read 2 more answers

In a square, one pair of opposite sides was made 50% longer and the other pair of opposite sides was made 50% shorter

omeli [17]

Sorry, maybe is too late for you, But the answer is:
For example: 4 x 4=16cm² if <span> one pair of opposite sides was made 50% longer and the other pair of opposite sides was made 50% shorter
It became: 6 x 2=12cm</span>²
12/16=0.75
So, the 75% <span>of the square’s area is the area of the new rectangle.</span>

8 0

3 years ago