26.45+4.79+120.02-3.20
31.24+120.02-3.20
151.26-3.20
148.06
<span>A 276,543 . . . . . No. The nearest thousand is 277,000 .
B 276,479 . . . . . Yes.
C 275,424 . . . . . No. The nearest thousand is 275,000 .
D 275,289</span> . . . . . No. The nearest thousand is 275,000 .
Tan(60)/cos(45) = (sin(60)/cos(60))/cos(45)
= ((√3/2)/(1/2))/(√2/2) = (√3)/(√2/2) = (2√3)/(√2)
= ((2√3)(√2))/((√2)(√2)) = (2√6)/2 = √6
If any step doesn't makes sense, hmu!
The angle that is coterminal to 425° is the last one:
B = 425° + n*1,440°
<h3>Which measure is of an angle that is coterminal with a 425° angle?</h3>
By definition, for any angle A, we can say that an angle B is coterminal to A if:
B = A + n*360°
where n can be any integer.
So, from the given options, we need to see which one is a multiple of 360°.
Of the given options, the only that meets this condition is the last one:
B = 425° + n*1,440°
Where:
1,440°/360° = 4
Then we conclude that:
425° + n*1,440° is coterminal to 425°.
If you want to learn more about coterminal angles:
brainly.com/question/3286526
#SPJ4
Answer:
1.4 times as high
Step-by-step explanation:
Now here we have to compare the both final velocities by using the formula independent of time. As we know,
2aS = V(f)2 - V(i)2
Both balls are dropped from certain height, so V(i) = 0 m/s for both balls
So, 2aS = V(f)2 for both balls.
Now if we compare both velocities
Equation for first ball
2aS(1) = V(1)2
Equation for second ball
2aS(2) = V(2)2
Comparing both equations
2aS(1)/2aS(2) = V(1)2/V(2)2
as we now S(1) = 5m and S(2) = 10m, also acceleration a = g, because acceleration due to gravity is acting on the balls. So,
2(9.8) x 5/2(9.8) x 10 = V(1)2/V(2)2
after simplifying,
1/2 = V(1)2/V(2)2
V(2)^2 = 2V(1)^2
By taking square root on both sides,
V(2) = 1.4V(1)