$2\sin x\cos x=\cos x\\\\2\sin x\cos x-\cos x=0\\\\\cos x(2\sin x-1)=0\iff\ \cos x=0\ or\ 2\sin x-1=0\\\\\cos x=0\ if\ x=\dfrac{\pi}{2}+k\pi\\\\2\sin x-1=0\\\\2\sin x=1\\\\\sin x=\dfrac{1}{2}\ if\ x=\dfrac{\pi}{6}+2k\pi\ or\ x=\dfrac{5\pi}{6}+2k\pi\\\\Answer:\boxed{x=\dfrac{\pi}{2}+k\pi\ or\ x=\dfrac{\pi}{6}+2k\pi\ or\ x=\dfrac{5\pi}{6}+2k\pi\ /k\in\mathbb{Z}/}$

7 0

3 years ago

Can someone plz help me with this math problem.

galben [10]

For question one, you would need to use the formula pi r ^2. So first you have 4.5, which is your radius(stands for r). Then you square it, which is 20.25. Then, you multiply 20.25 with pi. Then for circumference, you do 2 x pi x the radius which is 4.5. So for area, you get 63.62. Then for circumference you get 28.27.

For question 2, it is basically the same steps but instead you a diameter, not a radius. To get the radius, divide by 2, which the radius will be 6. Then from there, the radius is 6, and you solve the question.

4 0

3 years ago

Read 2 more answers

Rewrite the equation correctly.

Nataliya [291]

Answer:

Step-by-step explanation:

2x - y = 5 <em>subtract 2x from both sides</em>

2x - 2x - y = -2x + 5

-y = -2x + 5 <em>change the signs</em>

y = 2x - 5

7 0

3 years ago