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Anestetic [448]
3 years ago
9

Find the angles marked with with letters​

Mathematics
2 answers:
Sever21 [200]3 years ago
7 0

Answer:

<A=67⁰

<B=22⁰

<C=42⁰

<h2>Mark me BRAINLIEST </h2>
Helga [31]3 years ago
6 0

Answer:

a = 67° , b = 26° , c = 69°

Step-by-step explanation:

a and 67° are corresponding angles and are congruent , so

a = 67°

------------------------------------------------------

The triangle on the left has 3 equal sides thus is equilateral with 3 congruent angles of 60°

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles , that is

b + 34° = 60° ( subtract 34° from both sides )

b = 26°

---------------------------------------------------

The vertex of the triangle and 42° are vertically opposite and are congruent

vertex = 42°

The triangle has 2 equal legs and is isosceles with 2 base angles congruent.

c = \frac{180-42}{2} = \frac{138}{2} = 69°

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2 years ago
Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions
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I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take z=y', so that z'=y'' and we're left with the ODE linear in z:

y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2

Now suppose y has a power series expansion

y=\displaystyle\sum_{n\ge0}a_nx^n
\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Then the ODE can be written as

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0

\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0

All the coefficients of the series vanish, and setting x=0 in the power series forms for y and y' tell us that y(0)=a_0 and y'(0)=a_1, so we get the recurrence

\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}

We can solve explicitly for a_n quite easily:

a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}

and so on. Continuing in this way we end up with

a_n=\dfrac{a_1}{n!}

so that the solution to the ODE is

y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x

We also require the solution to satisfy y(0)=a_0, which we can do easily by adding and subtracting a constant as needed:

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4 0
3 years ago
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