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slega [8]
3 years ago
5

What's -3+(0+9) ?? I need to know by tonight

Mathematics
2 answers:
Crank3 years ago
6 0
The answer is 6.
To do this, remember PEMDAS? 
You would first have to add what is in the parenthesis THEN do -3 + 9 to get 6. 
Andre45 [30]3 years ago
6 0
-3 because you would ignore 0 since u would replace whatever is infront of the + with a 1, 1 and 0 makes 0, 1 x 9 = 9. then you are left with -3(9) you divided -3 and get -3. -3 x -3 = 9
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Find all solutions in the interval [0, 2π). <br> 2 sin2x = sin x
tester [92]

Answer:

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Step-by-step explanation:

We will need the double angle identity \sin(2x)=2\sin(x)\cos(x).

Let's begin:

2\sin(2x)=\sin(x)

Use double angle identity mentioned on left hand side:

2\cdot 2\sin(x)\cos(x)=\sin(x)

Simplify a little bit on left side:

4\sin(x)\cos(x)=\sin(x)

Subtract \sin(x) on both sides:

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Factor left hand side:

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Set both factors equal to 0 because at least of them has to be 0 in order for the equation to be true:

\sin(x)=0 \text{ or } 4\cos(x)-1=0

The first is easy what angles \theta are y-coordinates on the unit circle 0. That happens at 0 and \pi on the given range of x (this x is not be confused with the x-coordinate).

Now let's look at the second equation:

4\cos(x)-1=0

Isolate \cos(x).

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4 \cos(x)=1

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\cos(x)=\frac{1}{4}

This is not as easy as finding on the unit circle.

We know \arccos( ) will render us a value between 0 and 2\pi.

So one solution on the given interval for x is x=\cos^{-1}(\frac{1}{4}).

We know cosine function is even.

So an equivalent equation is:

\cos(-x)=\frac{1}{4}

Apply \cos^{-1} to both sides:

-x=\cos^{-1}(\frac{1}{4})

Multiply both sides by -1:

x=-\cos^{-1}(\frac{1}{4})

This going to be negative in the 4th quadrant but if we wrap around the unit circle, 2\pi , we will get an answer between 0 and 2\pi.

So the solutions on the given interval are :

0

\pi

\cos^{-1}(\frac{1}{4})

-\cos^{-1}(\frac{1}{4})+2\pi

5 0
3 years ago
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