3 years ago

Which statement correctly describes how to find an equivalent fraction for 60

Mathematics

1 answer:

Dahasolnce [82]3 years ago

3 0

Answer:

C and E

Step-by-step explanation:

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The circle shown is centered at the origin and contains the point (-4, -2). What is the length of the diameter of the circle? Ro

trasher [3.6K]

Step-by-step explanation:

Equation for circle centered at origin:

x² + y² = r².

Therefore r² = (-4)² + (-2)² = 20, r = √20.

Hence d = 2r = 2√20 = 8.94 (2d.p.)

8 0

3 years ago

20 POINTS!!!

notka56 [123]

$\dfrac{2}{\sqrt3\cos x+\sin x}=\sec\left(\dfrac{\pi}{6}-x\right)\\\\\dfrac{2}{\sqrt3\cos x+\sin x}=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}\ \ \ \ \ (*)\\----------------\\\\\cos\left(\dfrac{\pi}{6}-x\right)\ \ \ \ |\text{use}\ \cos(x-y)=\cos x\cos y+\sin x\sin y\\\\=\cos\dfrac{\pi}{6}\cos x+\sin\dfrac{\pi}{6}\sin x=\dfrac{\sqrt3}{2}\cos x+\dfrac{1}{2}\sin x=\dfrac{\sqrt3\cos x+\sin x}{2}\\------------------------------$

$(*)\\R_s=\sec\left(\dfrac{\pi}{6}-x\right)=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}=\dfrac{1}{\dfrac{\sqrt3\cos x+\sin x}{2}}\\\\=\dfrac{2}{\sqrt3\cos x+\sin x}=L_s$