I would tell her that her prediction is inaccurate because it isn’t guaranteed that the coin will land on both sides an equal amount of times. It’s possible that the coin could land heads up more often than it could land tails up or the other way around.
Step-by-step explanation:
-w - 11 < or = w - 5
-2w - 11 < or = -5
-2w < or = -16
w > or = -8
Answer:
The set of solutions is ![\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12%5C%5C-7-r%5C%5Cr%5Cend%7Barray%7D%5Cright%5D%3A%20%5Ctext%7Br%20is%20a%20real%20number%7D%20%20%5C%7D)
Step-by-step explanation:
The augmented matrix of the system is ![\left[\begin{array}{ccccc}3&6&6&-9\\-2&-3&-3&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D3%266%266%26-9%5C%5C-2%26-3%26-3%263%5Cend%7Barray%7D%5Cright%5D)
.
We will use rows operations for find the echelon form of the matrix.
- In row 2 we subtract
from row 1. (R2- 2/3R1) and we obtain the matrix ![\left[\begin{array}{cccc}3&6&6&-9\\0&1&1&-7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%266%266%26-9%5C%5C0%261%261%26-7%5Cend%7Barray%7D%5Cright%5D)
- We multiply the row 1 by
.
Now we solve for the unknown variables:
The system has a free variable, the the system has infinite solutions and the set of solutions is
Answer:
6x + 2
Explanation:
4 + -2 = 2 and x + 3x + 2x = 6x
<span><u><em>The correct answer is: </em></u>
we cannot take an even root of a negative number in the real number system.
<u><em>Explanation</em></u><span><u><em>: </em></u>
A rational exponent is another way of writing a root. The denominator of the rational exponent will be the root, and the numerator will be the power of the number under the root. If the denominator is even, this means we are taking an even root, such as square root, 4th root, etc.
If the base is negative, this means we want an even root of a negative number.
Any even root can be broken down into a series of square roots, done one after another.
However, we cannot take the square root of a negative number using real numbers; this is because a square root is a number multiplied by itself to get the answer; a negative times a negative is a positive, as is a positive times a positive, so there is no way to multiply the same number twice and get a negative.</span></span>
,
,
then