Recall:
The
tan of the measure of an angle is the ratio of the opposite side to the adjacent side to that angle, that is :

.
Since this ratio is 3/y, we denote the opposite side, and adjacent side respectively by 3 and y.
(Technically we should write 3t and yt, but we try our luck as we see y in the second ratio too!)
Similarly,

.
The adjacent side is already denoted by y, so we denote the length of the hypotenuse by z.
Now the sides of the right triangle are complete.

Answer: A
let's multiply both sides in each equation by the LCD of all fractions in it, thus doing away with the denominator.
![\begin{cases} \cfrac{1}{2}x+\cfrac{1}{3}y&=7\\\\ \cfrac{1}{4}x+\cfrac{2}{3}y&=6 \end{cases}\implies \begin{cases} \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{6}}{6\left( \cfrac{1}{2}x+\cfrac{1}{3}y \right)=6(7)}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( \cfrac{1}{4}x+\cfrac{2}{3}y\right)=12(6)} \end{cases}\implies \begin{cases} 3x+2y=42\\ 3x+8y=72 \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20%5Ccfrac%7B1%7D%7B2%7Dx%2B%5Ccfrac%7B1%7D%7B3%7Dy%26%3D7%5C%5C%5C%5C%20%5Ccfrac%7B1%7D%7B4%7Dx%2B%5Ccfrac%7B2%7D%7B3%7Dy%26%3D6%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B6%7D%7D%7B6%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7Dx%2B%5Ccfrac%7B1%7D%7B3%7Dy%20%5Cright%29%3D6%287%29%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B12%7D%7D%7B12%5Cleft%28%20%5Ccfrac%7B1%7D%7B4%7Dx%2B%5Ccfrac%7B2%7D%7B3%7Dy%5Cright%29%3D12%286%29%7D%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%203x%2B2y%3D42%5C%5C%203x%2B8y%3D72%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \stackrel{\textit{using elimination}}{ \begin{array}{llll} 3x+2y=42&\times -1\implies &\begin{matrix} -3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~-2y=&-42\\ 3x+8y-72 &&~~\begin{matrix} 3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+8y=&72\\ \cline{3-4}\\ &&~\hfill 6y=&30 \end{array}} \\\\\\ y=\cfrac{30}{6}\implies \blacktriangleright y=5 \blacktriangleleft \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20elimination%7D%7D%7B%20%5Cbegin%7Barray%7D%7Bllll%7D%203x%2B2y%3D42%26%5Ctimes%20-1%5Cimplies%20%26%5Cbegin%7Bmatrix%7D%20-3x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~-2y%3D%26-42%5C%5C%203x%2B8y-72%20%26%26~~%5Cbegin%7Bmatrix%7D%203x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%2B8y%3D%2672%5C%5C%20%5Ccline%7B3-4%7D%5C%5C%20%26%26~%5Chfill%206y%3D%2630%20%5Cend%7Barray%7D%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B30%7D%7B6%7D%5Cimplies%20%5Cblacktriangleright%20y%3D5%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \stackrel{\textit{substituting \underline{y} on the 1st equation}~\hfill }{3x+2(5)=42\implies 3x+10=42}\implies 3x=32 \\\\\\ x=\cfrac{32}{3}\implies \blacktriangleright x=10\frac{2}{3} \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left(10\frac{2}{3}~~,~~5 \right)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20%5Cunderline%7By%7D%20on%20the%201st%20equation%7D~%5Chfill%20%7D%7B3x%2B2%285%29%3D42%5Cimplies%203x%2B10%3D42%7D%5Cimplies%203x%3D32%20%5C%5C%5C%5C%5C%5C%20x%3D%5Ccfrac%7B32%7D%7B3%7D%5Cimplies%20%5Cblacktriangleright%20x%3D10%5Cfrac%7B2%7D%7B3%7D%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cleft%2810%5Cfrac%7B2%7D%7B3%7D~~%2C~~5%20%5Cright%29~%5Chfill)
Answer:
8
Step-by-step explanation:
Two different approaches:
<u>Method 1</u>
Apply radical rule √(ab) = √a√b to simplify the radicals:
√98 = √(49 x 2) = √49√2 = 7√2
√50 = √(25 x 2) = √25√2 = 5√2
Therefore, (√98 - √50)² = (7√2 - 5√2)²
= (2√2)²
= 4 x 2
= 8
<u>Method 2</u>
Use the perfect square formula: (a - b)² = a² - 2ab + b²
where a = √98 and b = √50
So (√98 - √50)² = (√98)² - 2√98√50 + (√50)²
= 98 - 2√98√50 + 50
= 148 - 2√98√50
Apply radical rule √(ab) = √a√b to simplify radicals:
√98 = √(49 x 2) = √49√2 = 7√2
√50 = √(25 x 2) = √25√2 = 5√2
Therefore, 148 - 2√98√50 = 148 - (2 × 7√2 × 5√2)
= 148 - 140
= 8
Answer:
The probability of the spinner landing on A is 0.1111.
Step-by-step explanation:
The possible outcomes of a board game is that it can lead on any three of the regions, A, B and C.
The probability of all the outcomes of an event sums up to be 1.

According to the provided information:
P (B) = 3 P (A)
P (C) = 5 P (A)
Compute the probability of the spinner landing on A as follows:

Thus, the probability of the spinner landing on A is 0.1111.