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butalik [34]
3 years ago
6

Which completely describes the polygon ? equilateral equiangular regular congruent

Mathematics
1 answer:
VladimirAG [237]3 years ago
5 0
<h3>Answer: C) regular</h3>

Explanation:

Any regular polygon is both equilateral and equiangular.

Equilateral = all sides are equal

Equiangular = all angles are equal

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The total number of choices Jalen has is six
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3 years ago
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10.Which of the following values does not satisfy the inequality ?
kupik [55]

Answer:

The answer is -4

Step-by-step explanation:

-2x-6<=1

-2(-4)-6<=1

8-6<=1

2<=1 which is False

-2x-6<=1

-2(-3)-6<=1

6-6<=1

0<=1 which is True

-2x-6<=1

-2(-2)-6<=1

4-6<=1

-2<=1 which is True

-2x-6<=1

-2(-1)-6<=1

2-6<=1

-4<=1 which is True

7 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
in order to be invited to tutoring or extension activities a student's performance must be at least 20 points higher or lower th
Ad libitum [116K]

Answer:

p ≤ 85 or p ≥ 125

Step-by-step explanation:

8 0
3 years ago
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Help with 9 and 10 please?
JulsSmile [24]
9. Let x equal the number

Two less than a number is

x - 2

Is more than 15 is

> 15

Put them together for

x-2 > 15

Now solve

x-2 > 15
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10. Let x equal a number

Seven more than a number is

x+ 7

Is less than or equal to 27 is

<= 27

Put them together for

x + 7 <= 27

Now solve

x+ 7 <= 27
x <= 20
3 0
3 years ago
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