Answer:
The system of this equation is ( -1, 10) which is Option A.
Step-by-step explanation:
Solve for y in the first equation.
Replace all occurrences of yy with -5x+5 in each equation.
Solve for x in the first equation.
Replace all occurrences of x with -1 in each equation.
The solution to the system is the complete set of ordered pairs that are valid solutions.
(−1, 10)
<u>Hence</u><u>,</u><u> </u><u>the</u><u> </u><u>system</u><u> </u><u>of</u><u> </u><u>this</u><u> </u><u>equation</u><u> </u><u>is</u><u> </u><u>(</u><u>-1</u><u>,</u><u> </u><u>10</u><u>)</u><u>.</u><u> </u>
<u>Option</u><u> </u><u>A</u><u>.</u>
The product of the given two matrices comes out to be ![\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
Here we are given the 2 matrices as follows-
![\left[\begin{array}{ccc}7&-2\\-6&2\end{array}\right] \left[\begin{array}{ccc}1&1\\3&3.5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-2%5C%5C-6%262%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C3%263.5%5Cend%7Barray%7D%5Cright%5D)
To find the product of 2 matrices, the number of columns in the first matrix must be equal to the number of rows in the second matrix.
Here since both of the matrices are 2 × 2, their product is possible.
Now, to find the product, we need to multiply each element in the first row by each element of the 1st column of the second matrix and then find their sum. Similarly, we do this for all rows and columns.
Therefore,
![\left[\begin{array}{ccc}(7*1)+(-2*3)&(7*1)+(-2*3.5)\\(-6*1)+(2*3)&(-6*1)+(2*3.5)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%287%2A1%29%2B%28-2%2A3%29%26%287%2A1%29%2B%28-2%2A3.5%29%5C%5C%28-6%2A1%29%2B%282%2A3%29%26%28-6%2A1%29%2B%282%2A3.5%29%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}(7)+(-6)&(7)+(-7)\\(-6)+(6)&(-6)+(7)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%287%29%2B%28-6%29%26%287%29%2B%28-7%29%5C%5C%28-6%29%2B%286%29%26%28-6%29%2B%287%29%5Cend%7Barray%7D%5Cright%5D)
=
Thus, the product of the given two matrices comes out to be
Learn more about matrices here-
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Answer:
C. 
Step-by-step explanation:
=
so to not get a 2, the experimental probability would be or C
Answer:
(A) 165
(B) 330
Step-by-step explanation:
Total number of students in the class = 11
(A) How many different combinations of 3 selected students can he create?
ⁿCₓ = n! ÷ [(n - x)! x!]
n = 11, x = 3
11! / [8! 3!] = 165
<em>HINT: 8! or 8 factorial represents [8x7x6x5x4x3x2x1]</em>
(B) How many different combinations of 4 selected students can he create?
ⁿCₓ = n! ÷ [(n - x)! x!]
n = 11, x = 4
11! / [7! 4!] = 330
<em>Same hint applies here, for all numbers with the factorial sign.</em>
Answer: 7
Step-by-step explanation: there are 4 full circles so that equals 4. 1+1=2, 1/8+7/8=1. 4+2+1=7
