So, first we multiply the fraction by using the formula a/b times c/d= a times c/b times d
=(y^2-16) times 5y/2y(y-4)
Now, we cancel the common factor y
=(y^2-16) times 5/2(y-4)
Now, we factor 5(y^2-16)
We factor (y^2-16) first
y^2-16
Rewrite 16 as 4^2
y^2-4^2
Now, apply the formula x^2-y^2=(x+y)(x-y)
=y^2-4^2=(y+4)(y-4)
=5(y+4)(y-4)
=5(y+4)(y-4)/2(y-4)
Cancel the common factor y-4
=5(y+4)/2
Answer: 5(y+4)/2
let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B6%7D%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B6%5E2-5%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B36-25%7D%5Cimplies%20%5Cpm%20%5Csqrt%7B11%7D%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
<span>The difference between the larger result and the smaller result was 10.
Suppose the 4 digits numbers are abcd and pqrs
Here, 1st Number is = 1000*a + 100*b + 10*c + d
and 2nd</span> Number is = 1000*ap + 100*q + 10*r + s
Then, since Kent made an error if 1 in tens columns and Harold added it correct, so the difference will be of 10 points.
