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Mathematics

1 answer:

Scorpion4ik [409]3 years ago

3 0

Angle 1 is congruent to angles 3, 5, and/or 7

Angle 2 is congruent to angles 4, 6, and/or 8

Angle 5 is congruent to angles 7, 3 and/or 1

Angle 6 is congruent to angles 8, 4, and/or 2

Any of these answers could work for the blanks.

Angles 1 and 3, 2 and 4, 5 and 7, and angles 6 and 8 are congruent because they are vertical angles. They have the same vertex. Not all of these are congruent to each other if this doesn’t make sense. It’s only 1 is congruent to 3, 2 congruent to 4, etc.

Then you have your corresponding angles. These are ones like angles 2 and 6, then 1 and 5. You can also have 8 and 4, or 7 and 3 as corresponding angles

Transversal angles are different. This would be like angles 3 and 4, or 1 and 2. They are not always congruent. The only time they will be congruent is if they are both 90°. Transversal angles are essentially supplementary angles on the transversal line (the line that intersects through the set of parallel lines)

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horrorfan [7]

Answer:

D. 40%

Step-by-step explanation:

15/6=40% AKA 4/10

8 0

2 years ago

Read 2 more answers

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

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$