Let’s call the speed of the slower car S, then the speed of the other is S+10mph.
At 5pm they have been travelling for 3 hours. The slower car travels a distance 3S and the faster one 3(S+10).
But the two distances must add up to 240 miles so 3S+3(S+10)=240, 3S+3S+30=240, 6S=210, S=35 mph. The faster car’s speed is 45mph. We can see that 3S is the same distance as 3x, so x=S=35 mph, and the distance the faster car travels is 3×45=135 miles.
Answer:
cot∅ = (-2√30)/7.
Step-by-step explanation:
Given the value of csc∅ = -13/7 and ∅ is in quad III.
We know y = r sin∅ and r > 0. So csc∅ = r/y = -13/7 = 13/(-7).
It means y = -7, r = 13.
We know x² + y² = r².
x² = r² - y²
x² = (13)² - (-7)² = 169 - 49 = 120.
x = √120 = 2√30.
we know cot∅ = x/y = (2√30)/(-7) = (-2√30)/7.
Hence, cot∅ = (-2√30)/7.
Just sayin' you shouldn't use up all ur points on one question
If i simplify number 6 it would be -55/63
