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masya89 [10]
3 years ago
5

Whats is the answer to this question ?

Mathematics
1 answer:
Lilit [14]3 years ago
6 0

Answer:

2 and 1

Step-by-step explanation:

adjacent means next t, and angles 2 and 1 are next to each other

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What is the value of n in the equation below?<br><br>A:-20<br>B:-9<br>C:9<br>D:20
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12^9/12^5 = 12^4

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Given points A (0,0), B (4,0), C (0,2), D (0,0), E (2,0), and F (0,1), which of the following proves that
damaskus [11]

Similar triangles may or may not have equal side lengths

Triangles ABC and DEF are similar by SSS theorem

<h3>How to determine the similarity statement</h3>

The coordinates of the two triangles are:

A (0,0), B (4,0), C (0,2), D (0,0), E (2,0), and F (0,1)

Calculate the lengths of both triangles using the following distance formula

d = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2}

For triangle ABC, we have:

AB = \sqrt{(0 -4)^2 + (0 -0)^2} = 4

BC = \sqrt{(4 -0)^2 + (0 -2)^2} = 2\sqrt 5

CA = \sqrt{(0 -0)^2 + (0 -2)^2} = 2

For triangle DEF, we have:

DE = \sqrt{(0 -2)^2 + (0 -0)^2} = 2

EF = \sqrt{(2 -0)^2 + (0 -1)^2} = \sqrt 5

FD = \sqrt{(0 -0)^2 + (1 -0)^2} = 1

Divide corresponding sides of the triangles to calculate the scale factor k

k = \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}

This gives

k = \frac{4}{2} = \frac{2\sqrt 5}{\sqrt 5} = \frac{2}{1}

Evaluate the quotients

k = 2 = 2 = 2

Since the quotients are equal, then the triangles are similar

Hence, triangles ABC and DEF are similar by SSS theorem

Read more about similar triangles at:

brainly.com/question/14285697

3 0
2 years ago
How to know if a function is periodic without graphing it ?
zhenek [66]
A function f(t) is periodic if there is some constant k such that f(t+k)=f(k) for all t in the domain of f(t). Then k is the "period" of f(t).

Example:

If f(x)=\sin x, then we have \sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x, and so \sin x is periodic with period 2\pi.

It gets a bit more complicated for a function like yours. We're looking for k such that

\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5

Expanding on the left, you have

\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2

and

1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5

It follows that the following must be satisfied:

\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}

The first two equations are satisfied whenever k\in\{0,\pm4,\pm8,\ldots\}, or more generally, when k=4n and n\in\mathbb Z (i.e. any multiple of 4).

The second two are satisfied whenever k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}, and more generally when k=\dfrac{10n}7 with n\in\mathbb Z (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when k is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2

\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5

More generally, it can be shown that

f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))

is periodic with period \mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n).
4 0
3 years ago
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