So, (weekly) she receives 1036.00.
The question is asking to find out the (annual) salary which is a year.
There are about 52 weeks in a year.
To find the amount being payed annually we multiply 1036 by 52
=53,872.
Answer:
Step-by-step explanation:
Sum
-2d + 1 + 6d + 4
Solving like terms
4d + 5
Difference
-2d + 1 - 6d - 4
-8d - 3
Answer:
(1,0)
Step-by-step explanation:
Absolute value is represented by using these bars → ║.
║= number of units from 0.
This equation says that whatever number is in the absolute value bars , x , is greater than y.
139, 149, 159, 169, 179, 189, 199, 209, 219, 229, 239, 249, 259.
The common difference is 13.
Let n = 52
Let d = common difference
a_52 = 139 + (52 - 1)(13)
a_52 = 139 + (51)(13)
a_52 = 139 + 663
a_52 = 802
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)
