∠R is congruent to ∠G
The reason: the right angle's orientation gives a reference for which angle, if the triangle is congruent, is congruent to which angle.
OR
You can follow the letters. If the triangle is labeled with the letters or numbers, it is 90% of the time going to be in order of the alphabet or the numerical sequence.
ΔGHI ≅ ΔRST
Notice how those letters are in order? The ∠G is congruent to ∠R.
Hope I could help you out! God bless!
#1). The length of the whole circumference is (π · D) = 12π .
120° is 1/3 of 360°, so arc-AB is 1/3 of the whole circumference.
#2). 60° is 1/6 of 360°.
So arc-CD is 1/6 of the whole circumference (which is given).
#3). The circumference of the circle is (π·D) = 20π .
70° is (70°/360°) of the whole circle, so arc-XY
is (70/360) of the whole circumference.
#4). Sorry, I don't remember anything right now about
angles between chords.
I hope what I've given you so far is worth 5 points.
Answer: 2.5 // 2 1/2
Step-by-step explanation:
1/2 + 1/2 = 1
3/4 + 3/4 = 6/4 = 1 1/2
1 + 1 1/2 = 2 1/2
2 1/2 = 2.5
Answer:
Um i think it would be
does not exist
Step-by-step explanation:
sorry if wrong im a little bad at that
Answer:
D) ![\left[\begin{array}{c}\frac{5}{4}\\-\frac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%5Cfrac%7B5%7D%7B4%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
For matrix ![\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
the inverse matrix is the transpose of the cofactor matrix, divided by the determinant: ![\dfrac{1}{ad-bc}\left[\begin{array}{cc}d&-b\\-c&a\end{array}\right]](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bad-bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dd%26-b%5C%5C-c%26a%5Cend%7Barray%7D%5Cright%5D)
Your inverse matrix is: ![\dfrac{1}{2(-3)-(1)(2)}\left[\begin{array}{cc}-3&-1\\-2&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%28-3%29-%281%29%282%29%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%26-1%5C%5C-2%262%5Cend%7Barray%7D%5Cright%5D)
so the solution is ...
![\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}\frac{3}{8}&\frac{1}{8}\\\frac{1}{4}&-\frac{1}{4}\end{array}\right] \cdot\left[\begin{array}{c}2\\4\end{array}\right] =\left[\begin{array}{c}\frac{5}{4}\\-\frac{1}{2}\end{array}\right] \qquad\text{matches selection D}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B3%7D%7B8%7D%26%5Cfrac%7B1%7D%7B8%7D%5C%5C%5Cfrac%7B1%7D%7B4%7D%26-%5Cfrac%7B1%7D%7B4%7D%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C4%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%5Cfrac%7B5%7D%7B4%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%5Cqquad%5Ctext%7Bmatches%20selection%20D%7D)
