Answer:
See explanation
Step-by-step explanation:
cos(θ)= 18/30 = 3/5
sin(θ)= 24/30 = 4/5
tan(θ)= 24/18 = 4/3
cos(ϕ)= 24/30 = 4/5
sin(ϕ)= 18/30 = 3/5
tan(ϕ)= 18/24 = 3/4
Let point (x, y) be any point on the graph, than the distance between (x, y) and the focus (3, 6) is sqrt((x - 3)^2 + (y - 6)^2) and the distance between (x, y) and the directrix, y = 4 is |y - 4|
Thus sqrt((x - 3)^2 + (y - 6)^2) = |y - 4|
(x - 3)^2 + (y - 6)^2 = (y - 4)^2
x^2 - 6x + 9 + y^2 - 12y + 36 = y^2 - 8y + 16
x^2 - 6x + 29 = -8y + 12y = 4y
(x - 3)^2 + 20 = 4y
y = 1/4(x - 3)^2 + 5
Required answer is f(x) = one fourth (x - 3)^2 + 5
Answer:
Weight of wastage=179.775kg
weight of rice cultivated= 585.225 kg
percentage of rice cultivated=76.5%
Step-by-step explanation:
Area of land=100 square meters
Cultivated rice=765kg
Wastage=23.5%
1) Weight of the wastage=23.5% of 765kg
=23.5/100 × 765
=17977.5 / 100
=179.775 kg
2) Weight and percentage of rice cultivated.
weight of rice cultivated = 765 kg - 179. 775 kg
= 585.225 kg
percentage of rice cultivated = 100 - 23.5
= 76.5%
3) if area is increased 40 times in size
New area=1000 square meters × 40
=40,000 square meters
Cultivated rice= 765kg × 40
=30,600 kg
Cultivated rice excluding wastage=585.225 kg × 40
=23,409 kg
The order does not matter (A,B,C same as B,C,A) so it's a combination 28 choose 5.
<span>Let makes the table neater. event D= 100m, even E=800m
(Event D) (Event E)
A won the race (Event A) 30 50
B won the race (Event B) 60 25
A and B tied (Event T) 10 25</span>
B did not win the race today. What length of the race was it more likely to have been? The question is asking in what event did the chance of B did not win(lose + ties) is more likely. Then, you need to determine the chance for B did not win <span>in both race.
Chance of B </span>did not win <span>100-m= 30+10/10=40%
</span>Chance of B did not win 800-m= 50+25/10=75%
The difference would be 75%-40%=35% more likely to lose 800-m race
