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scoundrel [369]
2 years ago
11

In the figure below AC And bd are diameters of circle P what is the arc measure a Bc in degrees

Mathematics
2 answers:
UNO [17]2 years ago
7 0

Answer:

321

Step-by-step explanation:

Setler79 [48]2 years ago
7 0

Answer: 155

Step-by-step explanation:

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Can you help me with this plz
Salsk061 [2.6K]

Hello there

Please check attached image for answer.

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>.</em><em>.</em><em>.</em>

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3 0
2 years ago
PLESE HELP WITH ANSWER. rewrite the function in the given form
kirza4 [7]

s hard and too long I'm only of class 13

5 0
3 years ago
What is the order of the steps to solve for x in the equation 3x-6x-7=29
dexar [7]
First, you should combine the x terms:

-3x - 7 = 29

Then, get rid of the negative 7, by adding 7 to both sides.

-3x = 22

Finally divide 22 by -3, to solve for x.

x = -22/3 or -7.33
7 0
3 years ago
Read 2 more answers
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
Find the gradient of the line<br> 2y - 8x = -2
natima [27]

answer:

gradient is 4

Step-by-step explanation:

gradient is another way of saying slope

slope in the math is often used to show the speed of something

it's the m in y= mx+b

make the equation look like

y = mx + b

2y - 8x = -2

2y = 8x - 2

y= (8x-2)/2

y=4x-1

gradient is the number next to x

answer : gradient is 4

8 0
1 year ago
Read 2 more answers
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