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s344n2d4d5 [400]
3 years ago
5

In the figure, a||b , and both lines are intersected by transversal t. Complete the statements to prove that ∠2 and ∠8 are suppl

ementary angles.
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m∠2 = m∠7 ( )
m∠7 + m∠8 = 180° ( )

Mathematics
1 answer:
agasfer [191]3 years ago
8 0

Answer:

Step-by-step explanation:

m∠2 ≅ m∠7 [Alternate exterior angles]

m∠7 + m∠8 = 180° [Linear pair theorem]

m∠2 + m∠8 = 180° [Substitution property]

∠2 and ∠8 are supplementary [Definition of supplementary angles]

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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
Solve the equation.<br> 2x – 3 + 4x = 21<br> x=-12<br> x= -6<br> x = 4<br> x = 6
WARRIOR [948]

Answer:

x = 4

Step-by-step explanation:

2x-3+4x=21

Isolate the variable by dividing each side by the factors that don't contain the variable

3 0
3 years ago
Read 2 more answers
You are charged 6.5% tax on a 42% purchase. Find the amount of tax
Rus_ich [418]
Assuming you means 6.5% tax of a $42 purchase

0.065*42=2.73

tax is $2.73
7 0
3 years ago
From a point A that is 8.20 m above level ground, the angle of elevation of the top of a building s 31 deg 20 min and the angle
Mariulka [41]

Answer:

30.11 meters ( approx )

Step-by-step explanation:

Let x be the distance of a point P ( lies on the building ) from the top of the building such that AP is perpendicular to the building and y be the distance of the building from point A, ( shown in the below diagram )

Given,

Point A is 8.20 m above level ground,

So, the height of the building = ( x + 8.20 ) meters,

Now, 1 degree = 60 minutes,

⇒ 1\text{ minute } =\frac{1}{60}\text{ degree }

20\text{ minutes }=\frac{20}{60}=\frac{1}{3}\text{ degree}

50\text{ minutes }=\frac{50}{60}=\frac{5}{6}\text{ degree}

By the below diagram,

tan ( 12^{\circ} 50') = \frac{8.20}{y}

tan(12+\frac{5}{6})^{\circ}=\frac{8.20}{y}

tan (\frac{77}{6})^{\circ}=\frac{8.20}{y}

\implies y=\frac{8.20}{tan (\frac{77}{6})^{\circ}}

Now, again by the below diagram,

tan (31^{\circ}20')=\frac{x}{y}

tan(31+\frac{1}{3})=\frac{x}{y}

\implies x=y\times tan(\frac{94}{3})=\frac{8.20}{tan (\frac{77}{6})^{\circ}}\times tan(\frac{94}{3})^{\circ}=21.9142943216\approx 21.91

Hence, the height of the building = x + 8.20 = 30.11 meters (approx)

8 0
3 years ago
If a =-2 and b= 2 what will be the value of a^b
raketka [301]
Write out the equation;

y=a^b 
y=(-2)^2 
y=(-2)(-2) 
y=4 

Therefore, the value of a^b is 4 

Hope I helped :) 
7 0
3 years ago
Read 2 more answers
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