I think is inside and then outside. Not sure
Hope this helps!
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)
1 yard = 3 feet
Divide 34 by 3, you should get 11 and 1/3.
So the classroom is 11 1/3 yards long, but 1/3 of a yard just so happens to be one foot, as 3 feet is 1 yard.
That means that the classroom is 11 yards and 1 foot long, or answer choice B.
Hope this helps! :)
Answer:
B
Step-by-step explanation:
Angles a and b are complementary angles because they both add up to 90 degrees, which fulfills the definition of complementary.
Answer:
75 ≤ length < 85
Step-by-step explanation:
You know a number 5 or greater will round up to the next higher number in the number place to its left. Similarly, numbers less than 5 round down (get replaced by zero).
This means that 75 will round up to 80. and 84.9999... will round down to 80.
a) The lower bound on length is 75. The length must be at least 75.
b) The upper bound on length is 85. The length must be less than 85.
