<em>Paco could buy 1/2 a yard, 5/8 yard, 3/4 yard, 7/8 yard, 1 yard, 1 1/8, so fourth since these fractions are all greater than 1. </em>
19-13p=-17p-5
+5 +5
24-13p=-17p
+13p +13p
24=-4p
/-4 /-4
-6=p
Since congruence and similarity for polygons is posted in a specific order, we know that angle A=angle F, side A=side F, die B=side G, and so on. Therefore, we have that angle C=angle H = 9x-7=5x+13. Subtracting 5x from both sides to separate the x using the subtraction property of equality, we get 4x-7=13. Next, we can add 7 to both sides to get 4x=20, and divide both sides by 4 to isolate the x and get x=5. Plugging that into angle C or angle H, we get that 9x-7=9*5-7=45-7=38=5*5+13.
Feel free to ask further questions!
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)
