Answer:
(0, 1).
Method 1 (Substitution):
Substituting our two y's, we get the following:
Thus, the only set of solutions is (0, 1). A quick sketch (either by hand or on Desmos) can confirm this.
Method 2 (Elimination):
We have two equations. We'll let the top one be equation 1 and the bottom one be equation 2. Eliminating as many variables as we can, we subtract (2) from (1) to get:
0 = 3x => x = 0.
So the only set of solutions is (0, 1).
Method 3 (Gaussian elimination):
We can place this in an augmented matrix and row reduce.
![\left[\begin{array}{cccc}1&5&1 & 1\\1&2&1 & 1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%265%261%20%26%201%5C%5C1%262%261%20%26%201%5Cend%7Barray%7D%5Cright%5D)
Row reducing this gives us:
![\left[\begin{array}{cccc}1&5&1 & 1\\0&3&0 & 0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%265%261%20%26%201%5C%5C0%263%260%20%26%200%5Cend%7Barray%7D%5Cright%5D)
This tells us that the only solution for x is x = 0 (since we read this as "3x = 0") and thus, the only solution we get is (0, 1).
Answer:
t > 20
Step-by-step explanation:
14 - 0.6t > 2
14 - 14 - 0.6t > 2 - 14
- 0.6t > - 12
- 0.6t ÷ - 0.6 > - 12 ÷ - 0.6
t > 20
Answer:
The least tile would decrease
69- third quartile
66- median
Step-by-step explanation:
The first dot is the minimum, first line is the first quartile, second line is the median, third line is the third quartile, and the last dot is the maximum.
You already have all you need. Just put in it the slope-intercept form: y=mx+b where m is the slope and b is the y-intercept so
y=7x-4
Answer: -iSimplify the expression using the definition of an imaginary number
i
=
√
−
1
.
−
i
Step-by-step explanation: Simplify the expression using the definition of an imaginary number i
=
√
−
1
.
−
i
