76/4=19 So with 19 questions correct and 10 incorrect, 29 questions were on the test. p.s. If I misunderstood and this isn't answering all your questions just comment or pm.

8 0

2 years ago

Read 2 more answers

Find all values of x in the interval [0,2π] that satisfy the equation sin2x=cosx. enter your answers as a list of numbers separa

yuradex [85]

$\sin(2x) = \cos(x) \\ 2 \sin(x) \cos(x) = \cos(x) \\ \cos(x) (2 \sin(x) - 1) = 0$
so
$\cos(x) = 0 > > > x = \frac{\pi}{2} + k\pi \\ \sin(x) = \frac{1}{2} > > > x = \frac{\pi}{6} + 2k\pi \\ or \: x = \frac{5\pi}{6} + 2k\pi$
$x = \frac{\pi}{2} or \frac{3\pi}{2} \\ x = \frac{\pi}{6} or \frac{5\pi}{6}$

6 0

3 years ago

Please help. Will mark Brainliest

mario62 [17]

First is B, since 1 and 3 are corresponding angles, they must be equal

Next is I, 3 and 5 must be equal since they are vertically opposite

Last is transitive property
Since 1 = 3 and 3 = 5, then 1 must equal 5

7 0

2 years ago