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2 years ago

Find the difference. m+2/(m-4)^2 - m-2/(m-4)^2

Mathematics

1 answer:

jeka57 [31]2 years ago

6 0

Answer:

4/(m-4)^2

Step-by-step explanation:

=(m+2)/(m-4)^2-(m-2)/(m-4)^2

=m+2-m+2/(m-4)^2

= 4/(m-4)^2

This is the correct answer.

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In ΔDEF, EF = 15, FD = 13, and DE = 4. Which list has the angles of ΔDEF in order from smallest to largest?

Rudik [331]

Answer:

< F , < E, < D

Step-by-step explanation:

ΔDEF

Arrange the sides from smallest to largest

DE = 4, FD = 13, and EF = 15.

Largest side opposes the largest angle, medium side opposes the medium angle, smallest side opposes the smallest angle.

the angles of ΔDEF in order from smallest to largest

< F , < E, < D

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1 year ago

Which of the following are properties of right triangles?

sineoko [7]

The properties that apply true are:

1–as it implies the Pythagorean theorem; a^2+b^2=c^2

3–as it implies the basic rules of visualized geometry.

4–as the greater side(hypotenuse) is opposite the right angle.

NOT 2, as the accrue angles ARE complimentary.

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2 years ago

0.48 divided by 0.06 with shown work please help!

Valentin [98]

Answer: 8

Step-by-step explanation:

2 To convert to a percent, multiply the decimal by 100.

8\times 100=800

8×100=800

800

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2 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

Please help me with this I will name branlist

otez555 [7]

I wish I knew but same profile pic pog

8 0

2 years ago

Read 2 more answers

Find the difference. m+2/(m-4)^2 - m-2/(m-4)^2​

Find the difference. m+2/(m-4)^2 - m-2/(m-4)^2