For the same reasons as the last problem...
s(n)=20(1-(1/2)^n)/(1-1/2)
s(n)=40(1-(1/2)^n)
But this time we are not starting at one so we must subtract the sum of s(2) from the sum of s(12) so:
40((1-(1/2)^12-(1-(1/2)^2)
40(0.249755859375)
≈9.99 (to nearest one-thousandth)
Right triangle, so one angle = 90 and sum of the other 2 angles = 90
so
71x - 1 + 20x = 90
91x = 91
x = 1
<A = 20(1) = 20
answer
<A = 20 degrees
Answer:
The outcome of guessing the correct answers to a 50 question test is 12.5.
Step-by-step explanation:
Let us first consider one question.
Since it has 4 options so we can say that;
Total number of possible outcomes = 4
Since 1 answer is correct so we can say that;
Favorable outcome = 1
Now we can say that;
Probability for choosing correct answer can be calculated by Favorable outcome divided by Total number of possible outcomes.
framing in equation form we get;
![P(Correct\ Answer) = \frac{1}{4}](https://tex.z-dn.net/?f=P%28Correct%5C%20Answer%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D)
Now Given:
Total number of question = 50
So we can say that;
Total number of correct answer out of 50 is equal to Probability for choosing correct answer multiplied by Total number of question.
framing in equation form we get;
Guessing number of correct answer out of 50 = ![\frac14 \times 50 = 12.5](https://tex.z-dn.net/?f=%5Cfrac14%20%5Ctimes%2050%20%3D%2012.5)
Hence The outcome of guessing the correct answers to a 50 question test is 12.5.
The answer for B is 2 and 1/4 is the most common figure size
Answer:
![\large\boxed{x=0\ and\ x=\pi}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7Bx%3D0%5C%20and%5C%20x%3D%5Cpi%7D)
Step-by-step explanation:
![\tan^2x\sec^2x+2\sec^2x-\tan^2x=2\\\\\text{Use}\ \tan x=\dfrac{\sin x}{\cos x},\ \sec x=\dfrac{1}{\cos x}:\\\\\left(\dfrac{\sin x}{\cos x}\right)^2\left(\dfrac{1}{\cos x}\right)^2+2\left(\dfrac{1}{\cos x}\right)^2-\left(\dfrac{\sin x}{\cos x}\right)^2=2\\\\\left(\dfrac{\sin^2x}{\cos^2x}\right)\left(\dfrac{1}{\cos^2x}\right)+\dfrac{2}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=2](https://tex.z-dn.net/?f=%5Ctan%5E2x%5Csec%5E2x%2B2%5Csec%5E2x-%5Ctan%5E2x%3D2%5C%5C%5C%5C%5Ctext%7BUse%7D%5C%20%5Ctan%20x%3D%5Cdfrac%7B%5Csin%20x%7D%7B%5Ccos%20x%7D%2C%5C%20%5Csec%20x%3D%5Cdfrac%7B1%7D%7B%5Ccos%20x%7D%3A%5C%5C%5C%5C%5Cleft%28%5Cdfrac%7B%5Csin%20x%7D%7B%5Ccos%20x%7D%5Cright%29%5E2%5Cleft%28%5Cdfrac%7B1%7D%7B%5Ccos%20x%7D%5Cright%29%5E2%2B2%5Cleft%28%5Cdfrac%7B1%7D%7B%5Ccos%20x%7D%5Cright%29%5E2-%5Cleft%28%5Cdfrac%7B%5Csin%20x%7D%7B%5Ccos%20x%7D%5Cright%29%5E2%3D2%5C%5C%5C%5C%5Cleft%28%5Cdfrac%7B%5Csin%5E2x%7D%7B%5Ccos%5E2x%7D%5Cright%29%5Cleft%28%5Cdfrac%7B1%7D%7B%5Ccos%5E2x%7D%5Cright%29%2B%5Cdfrac%7B2%7D%7B%5Ccos%5E2x%7D-%5Cdfrac%7B%5Csin%5E2x%7D%7B%5Ccos%5E2x%7D%3D2)
![\dfrac{\sin^2x}{(\cos^2x)^2}+\dfrac{2-\sin^2x}{\cos^2x}=2\\\\\text{Use}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-(1-\cos^2x)}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-1+\cos^2x}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{1+\cos^2x}{\cos^2x}=2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csin%5E2x%7D%7B%28%5Ccos%5E2x%29%5E2%7D%2B%5Cdfrac%7B2-%5Csin%5E2x%7D%7B%5Ccos%5E2x%7D%3D2%5C%5C%5C%5C%5Ctext%7BUse%7D%5C%20%5Csin%5E2x%2B%5Ccos%5E2x%3D1%5Cto%5Csin%5E2x%3D1-%5Ccos%5E2x%5C%5C%5C%5C%5Cdfrac%7B1-%5Ccos%5E2x%7D%7B%28%5Ccos%5E2x%29%5E2%7D%2B%5Cdfrac%7B2-%281-%5Ccos%5E2x%29%7D%7B%5Ccos%5E2x%7D%3D2%5C%5C%5C%5C%5Cdfrac%7B1-%5Ccos%5E2x%7D%7B%28%5Ccos%5E2x%29%5E2%7D%2B%5Cdfrac%7B2-1%2B%5Ccos%5E2x%7D%7B%5Ccos%5E2x%7D%3D2%5C%5C%5C%5C%5Cdfrac%7B1-%5Ccos%5E2x%7D%7B%28%5Ccos%5E2x%29%5E2%7D%2B%5Cdfrac%7B1%2B%5Ccos%5E2x%7D%7B%5Ccos%5E2x%7D%3D2)
![\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{(1+\cos^2x)(\cos^2x)}{(\cos^2x)^2}=2\qquad\text{Use the distributive property}\\\\\dfrac{1-\cos^2x+\cos^2x+\cos^4x}{\cos^4x}=2\\\\\dfrac{1+\cos^4x}{\cos^4x}=2\qquad\text{multiply both sides by}\ \cos^4x\neq0\\\\1+\cos^4x=2\cos^4x\qquad\text{subtract}\ \cos^4x\ \text{from both sides}\\\\1=\cos^4x\iff \cos x=\pm\sqrt1\to\cos x=\pm1\\\\ x=k\pi\ for\ k\in\mathbb{Z}\\\\\text{On the interval}\ 0\leq x](https://tex.z-dn.net/?f=%5Cdfrac%7B1-%5Ccos%5E2x%7D%7B%28%5Ccos%5E2x%29%5E2%7D%2B%5Cdfrac%7B%281%2B%5Ccos%5E2x%29%28%5Ccos%5E2x%29%7D%7B%28%5Ccos%5E2x%29%5E2%7D%3D2%5Cqquad%5Ctext%7BUse%20the%20distributive%20property%7D%5C%5C%5C%5C%5Cdfrac%7B1-%5Ccos%5E2x%2B%5Ccos%5E2x%2B%5Ccos%5E4x%7D%7B%5Ccos%5E4x%7D%3D2%5C%5C%5C%5C%5Cdfrac%7B1%2B%5Ccos%5E4x%7D%7B%5Ccos%5E4x%7D%3D2%5Cqquad%5Ctext%7Bmultiply%20both%20sides%20by%7D%5C%20%5Ccos%5E4x%5Cneq0%5C%5C%5C%5C1%2B%5Ccos%5E4x%3D2%5Ccos%5E4x%5Cqquad%5Ctext%7Bsubtract%7D%5C%20%5Ccos%5E4x%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5C1%3D%5Ccos%5E4x%5Ciff%20%5Ccos%20x%3D%5Cpm%5Csqrt1%5Cto%5Ccos%20x%3D%5Cpm1%5C%5C%5C%5C%20x%3Dk%5Cpi%5C%20for%5C%20k%5Cin%5Cmathbb%7BZ%7D%5C%5C%5C%5C%5Ctext%7BOn%20the%20interval%7D%5C%200%5Cleq%20x%3C2%5Cpi%2C%5C%20%5Ctext%7Bthe%20solutions%20are%7D%5C%20x%3D0%5C%20%5Ctext%7Band%7D%5C%20x%3D%5Cpi.)