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2 years ago

Are these triangles similar triangles?

Mathematics

2 answers:

Eddi Din [679]2 years ago

5 0

Answer:

Yes, they are similar according to Angle Side Angle postulate.

Step-by-step explanation:

Zanzabum2 years ago

3 0

A. Angle side angle

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Do the following lengths satisfy the Pythagorean Theorem? Lengths =14, 50, 48.<br> Yes<br> No

seraphim [82]

Answer:

yes

Step-by-step explanation:

14² +48² = 196 +2304 = 2500 = 50²

The three lengths (14, 48, 50) are a Pythagorean triple.

<em>Additional comment</em>

The two smaller numbers correspond to side lengths. The larger one corresponds to the hypotenuse length.

We typically write the numbers of a Pythagorean triple in increasing order, but the order doesn't really matter. As sides of a triangle, the triangle formed is a right triangle, regardless of the order.

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1 year ago

Given the diagram below, what is cos 45

Mamont248 [21]

Hey

Your answer is pretty simple

1/√2
Don't forget to mark Brainliest :^)

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3 years ago

Which method would prove the quadrilateral is a parallelogram?

FrozenT [24]

Answer:

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1 year ago

Read 2 more answers

PLEASE HELP PICTURE ATTACHED

jeyben [28]

Answer: Its 48in

Step-by-step explanation:

Love you all. And good luck.

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3 years ago

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

2 years ago