Answer:
QH = 227.8 km ≅ 228 km
Step-by-step explanation:
∵ The bearing from H to P is 084°
∵ The bearing from P to Q is 210°
∵ The distance from H to P = 340 km
∵ The distance from P to Q = 160 km
∴ The angle between 340 and 160 = 360 - 210 - (180 - 84) = 54°
( 180 - 84) ⇒ interior supplementary
By using cos Rule:
(QH)² = (PH)² + (PQ)² - 2(PH)(PQ)cos∠HPQ
(QH)² = 340² + 160² - 2(340)(160)cos(54) = 51904.965
∴ QH = 227.8 km ≅ 228 km
AC is a tangent so by definition, it touches the circle at exactly one point (point C) and forms a right angle at the tangency point. So angle ACO is 90 degrees
The remaining angle OAC must be 45 degrees because we need to have all three angles add to 180
45+45+90 = 90+90 = 180
Alternatively you can solve algebraically like so
(angle OAC) + (angle OCA) + (angle COA) = 180
(angle OAC) + (90 degrees) + (45 degrees) = 180
(angle OAC) + 90+45 = 180
(angle OAC) + 135 = 180
(angle OAC) + 135 - 135 = 180 - 135
angle OAC = 45 degrees
Side Note: Triangle OCA is an isosceles right triangle. It is of the template 45-45-90.
The number of throws that the player make X follows binomial distribution.
The probability that the player makes x throws out of eleven throws is
The probability that the player makes at most nine out of eleven free throws
is
Correct choice is (B).
That doesn’t make any sense to me sorry
The required value of (g°h)(-3) is 5.
Step-by-step explanation:
Given,
g(x)= x-2 and h(x) = 4 - x
To find (g°h)(-3)
Now,
(g°h)(x) = g(h(x))
= g(4-x)
= (4-x)-2 = 2-x
So,
(g°h)(-3) = 2-(-3) = 5 [ putting x=-3]
