now let's recall that a full circle has 360°, so then 1140 = 360 + 360 + 360 + 60.
namely, 1140 is 3 full-go-around the circle and then it lands on the I Quadrant at 60°. Thus cos(1140°) = cos(60°).
what's that cosine of 60°? Check the picture below.
2, Please Brainiest answer<span />
Its all rhombus , square and rectangle with all the sides √50
Answer:
156
Step-by-step explanation:
209x18.50=3866.5
5700-3866.5=1833.5
1833.5÷11.75=156.04
<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>
