Try this option:
1 meter=1000 mm; 1 hour=60 min. From another side: 1 mm=0.001 m. and 1 min=1/60 h.
Using this rule:
620 mm/h=0.06*620 m/h=37.2 meters per hour.
answer: 0.06
Answer:
i dont knwo
Step-by-step explanation:
Answer:
6 pounds
Step-by-step explanation:
1/4 times 3 = 3/4
3/8 times 2: 6/8 = 3/4
1/2 times 4: 4/2 = 2
5/8 times 4: 20/8 = 2 1/2
3/4 + 3/4 = 6/4 = 1 1/2
2 + 2 1/2 + 1 1/2 = 6 pounds
That'd be true only if the value of "s" is the exact same one for both
namely if sec(s) = cos(s)
then solving for "s"
thus
![\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s) \\\\\\ \pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides} \\\\\\ cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s](https://tex.z-dn.net/?f=%5Cbf%20sec%28s%29%3Dcos%28s%29%5Cqquad%20but%5Cimplies%20sec%28%5Ctheta%29%3D%5Ccfrac%7B1%7D%7Bcos%28%5Ctheta%29%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%5Ccfrac%7B1%7D%7Bcos%28s%29%7D%3Dcos%28s%29%5Cimplies%201%3Dcos%5E2%28s%29%5Cimplies%20%5Cpm%20%5Csqrt%7B1%7D%3Dcos%28s%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%201%3Dcos%28s%29%5Cimpliedby%20%5Ctextit%7Bnow%20taking%20%7Dcos%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%28%5Cpm%201%29%3Dcos%5E%7B-1%7D%5Bcos%28s%29%5D%5Cimplies%20cos%5E%7B-1%7D%28%5Cpm%201%29%3D%5Cmeasuredangle%20s)
Answer:
<em>$4.75</em>
Step-by-step explanation:
(had) 8.25 + (deposited) 2.25 = (total) 10.50
10.50 - (withdrew) 5.75 = (total left) 4.75
