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:
1.33
Step-by-step explanation:
Given that:
Upper specification limit ( USL ) = 10
Lower specification limit ( LSL ) = 2
the standard deviation σ = 1
What is the Cp"
The Cp is the process capability ratio which can be expressed by the formula:
Let x be the number of minutes of 5.8*10^6 seconds.
Each minute is made of 60 seconds. Let's then use the rule of three:
60 seconds --> 1 minute
5.8*10^6 --> x minutes
x = (5.8*10^6)*1 / 60
x ≈ 96,666.66 minutes.
So 5.8*10^6 seconds is about 96,666.66 minutes.
Hope this Helps! :)
Subtract the two prices to find the difference, then divide the difference by the original price
.
880 - 640 = 240
240 / 880 =0. 2727 = 27.27% discount Round the answer as needed.
Answer: " r = a + b - s " .
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Step-by-step explanation:
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Given:
a + b = s + r ; Solve for: "r" ; that is; rewrite the equation in terms of: " r " .
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Subtract "s" from each side of the equation;
to isolate "r" on one side of the equation:
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→ a + b - s = s + r - s ;
→ to get:
→ a + b - s = r ;
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↔ " r = a + b - s " .
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→ which is the answer.
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Hope this helps!
Best wishes to you!
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