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natita [175]
3 years ago
6

How many b angles are in an obus triangle

Mathematics
1 answer:
Masteriza [31]3 years ago
6 0

Answer:

An obtuse triangle has 1 obtuse angle and 2 acute angles.

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-3(u+2)=5u-1+5(2u+1)
Mazyrski [523]

Answer:

u = -5/9

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS
  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define equation</u>

-3(u + 2) = 5u - 1 + 5(2u + 1)

<u>Step 2: Solve for </u><em><u>u</u></em>

  1. Distribute:                             -3u - 6 = 5u - 1 + 10u + 5
  2. Combine like terms:             -3u - 6 = 15u + 4
  3. Add 3u to both sides:          -6 = 18u + 4
  4. Subtract 4 on both sides:    -10 = 18u
  5. Divide 18 on both sides:      -10/18 = u
  6. Simplify:                                -5/9 = u
  7. Rewrite:                                 u = -5/9

<u>Step 3: Check</u>

<em>Plug in u into the original equation to verify it's a solution.</em>

  1. Substitute in <em>u</em>:                     -3(-5/9 + 2) = 5(-5/9) - 1 + 5(2(-5/9) + 1)
  2. Multiply:                                -3(-5/9 + 2) = -25/9 - 1 + 5(-10/9 + 1)
  3. Add:                                      -3(13/9) = -25/9 - 1 + 5(-1/9)
  4. Multiply:                                -13/3 = -25/9 - 1 - 5/9
  5. Subtract:                               -13/3 = -34/9 - 5/9
  6. Subtract:                               -13/3 = -13/3

Here we see that -13/3 does indeed equal -13/3.

∴ u = -5/9 is a solution of the equation.

3 0
3 years ago
Read 2 more answers
69/14 as a mixed number
Otrada [13]

Answer:

6914 is 4 with a remainder of 13

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Solve using Fourier series.
Olin [163]
With 2L=\pi, the Fourier series expansion of f(x) is

\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L
\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx

where the coefficients are obtained by computing

\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx
\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx

\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx
\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx

\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx
\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx

You should end up with

a_0=0
a_n=0
(both due to the fact that f(x) is odd)
b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)

Now the problem is that this expansion does not match the given one. As a matter of fact, since f(x) is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of f(x), which is to say we're actually considering the function

\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3

and enforcing a period of 2L=2\pi. Now, you should find that

\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)

The value of the sum can then be verified by choosing x=0, which gives

\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)
\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots

as required.
5 0
3 years ago
Set a = 3 and k = 1. Use the slider to manipulate M. What happens to the graph when M is negative?
ANEK [815]

Answer:

C (reflection across the x access)

Step-by-step explanation:

I got it right on edge

5 0
3 years ago
Can you help me please I need it and this graduate
yanalaym [24]
Here is how to work the problems and write the expressions.

8 0
3 years ago
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