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USPshnik [31]
3 years ago
6

Question 10 of the picture

Mathematics
1 answer:
cricket20 [7]3 years ago
6 0
This question is pretty hard, but not impossible.
The key to solving this is to spot triangles.

Since we want to prove that ∠XBP and ∠YBQ are equal, we need to know their relationship towards one another.
If we look at Δs PQB and XYB, we can note down their relationship.

Thus, the way to begin is to prove either similar or congruent triangles.
In Δs PQB and XYB,
∠QPB = ∠YXB (angles in the alternate segment standing on the same arc, PX, are equal)
Similarly, ∠XYB = ∠PQB ("  ^  ") note: this means see reasoning above

Since we proved two angles are equal, by definition, the third angle must be equal.
Thus, ∠PBQ = ∠PBY (by definition, they are equal)
Since the two angles share a common angle between them, namely the angle marked on the diagram, then this means ∠XBP must be equal to ∠YBQ, as required.
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HELP ME PLZZ I AM GIVING 30 POINTS FOR THIS!!
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Answer:

Option D

Step-by-step explanation:

because you have the ability to get 2/5 answers if it is supposed to be probable less than 3.

5 0
2 years ago
PLEASE ..... HELP THANK YOU
lara [203]
5x-10=3x+40
5x=3x+50
2x=50
x=25
5*25=125
125-10=115
angle AEB=115

Hope this helps :)
4 0
3 years ago
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Complete the statement with the choice that best
chubhunter [2.5K]

Answer:

Factors of the constant term

Step-by-step explanation:

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3 years ago
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Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7
Zolol [24]
f(x)=\begin{cases}1&\text{for }-7

The Fourier series expansion of f(x) is given by

\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7

where we have

a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx
a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)
a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2

The coefficients of the cosine series are

a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx
a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)
a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}
a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}

When n is even, the numerator vanishes, so we consider odd n, i.e. n=2k-1 for k\in\mathbb N, leaving us with

a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}

Meanwhile, the coefficients of the sine series are given by

b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx
b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)
b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}
b_n=\dfrac{7(-1)^{n+1}}{n\pi}

So the Fourier series expansion for f(x) is

f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7
3 0
3 years ago
Write each percent as a decimal a 80% B 3% C 12.5% D 125%​
Tema [17]

Answer:

a .8

b .03

c .125

d 125.0

Step-by-step explanation:

4 0
3 years ago
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