Answer:
The answers are 0, 1 and −2.
Step-by-step explanation:
Let α=arctan(2tan2x) and β=arcsin(3sin2x5+4cos2x).
sinβ2tanβ21+tan2β2tanβ21+tan2β23tanxtan2β2−(9+tan2x)tanβ2+3tanx(3tanβ2−tanx)(tanβ2tanx−3)tanβ2=3sin2x5+4cos2x=3(2tanx1+tan2x)5+4(1−tan2x1+tan2x)=3tanx9+tan2x=0=0=13tanxor3tanx
Note that x=α−12β.
tanx=tan(α−12β)=tanα−tanβ21+tanαtanβ2=2tan2x−13tanx1+2tan2x(13tanx)or2tan2x−3tanx1+2tan2x(3tanx)=tanx(6tanx−1)3+2tan3xor2tan3x−3tanx(1+6tanx)
So we have tanx=0, tan3x−3tanx+2=0 or 4tan3x+tan2x+3=0.
Solving, we have tanx=0, 1, −1 or −2.
Note that −1 should be rejected.
tanx=−1 is corresponding to tanβ2=3tanx. So tanβ2=−3, which is impossible as β∈[−π2,π2].
The answers are 0, 1 and −2.
Given the scale is 1 cm to 25 km.
We have to find the length of the trail from centimeter to the nearest tenth of kilometer.
Let's take the length of the trail be x km.
The length of the trail given 7.2 cm. We will have to find the length in km.
We have to set it up as a proportion. So we can write the proportion as,
We have to cross multiply first. We will get,
So we have got the required length of the train in kilometer.
The actual length of the trail = 180km.
Answer:
Actually the answer would be 120.
4.16×48=200
3/5×200=120.
120 people out of five play.
Answer:
45 it is 45 45 um yea I ben did thiss
