Answer:
![\left[\begin{array}{ccc}7\\4\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%5C%5C4%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
The answer is a single-column matrix (7,4,2)
Step-by-step explanation:
In such multiplication of matrices, you have to proceed by multiplying each ROW of the first matrix by the COLUMN of the second matrix. So,
![\left[\begin{array}{ccc}3&6&1\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (3 * 2) + (6 * 0) + (1 * 1) = 6 + 0 + 1 = 7](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%266%261%5Cend%7Barray%7D%5Cright%5D%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C0%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%283%20%2A%202%29%20%2B%20%286%20%2A%200%29%20%2B%20%281%20%2A%201%29%20%3D%206%20%2B%200%20%2B%201%20%3D%207)
then...
![\left[\begin{array}{ccc}2&4&0\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (2 * 2) + (4 * 0) + (0 * 1) = 4 + 0 + 0 = 4](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%264%260%5Cend%7Barray%7D%5Cright%5D%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C0%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%282%20%2A%202%29%20%2B%20%284%20%2A%200%29%20%2B%20%280%20%2A%201%29%20%3D%204%20%2B%200%20%2B%200%20%3D%204)
and
![\left[\begin{array}{ccc}0&6&2\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (0 * 2) + (6 * 0) + (2 * 1) = 0 + 0 + 2= 2](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%266%262%5Cend%7Barray%7D%5Cright%5D%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C0%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%280%20%2A%202%29%20%2B%20%286%20%2A%200%29%20%2B%20%282%20%2A%201%29%20%3D%200%20%2B%200%20%2B%202%3D%202)
I hope it helps.
Answer:

is equivalent to
.
This is because 
The rule is, in order to add fractions, your denominators must be equivalent. In other words, you must have the same denominator for both fractions.
<em>Now what do we do? We add!</em>

This is because 4 + 1 = 5.
Your answer is
.
I hope this helps!
<h2><u>
PLEASE MARK BRAINLIEST!</u></h2>
Here you would want to use the sine function, since it’s making a triangle with the ground, and length of the string, the height to the kite. This can be found by 65sin(70°) which is ≈ 61.08 meters
ok so i can’t answer all of them BUT!! if a=a: infinitely many, x=a: one solution, a=k: zero solutions. hope this helps!
All the points that are 6 units from (-1, 1) are those on the circle
(x+1)^2 +(y-1)^2 = 36
For y=0, the two points of interest satisfy
(x+1)^2 +1 = 36
(x+1)^2 = 35 . . . . . . subtract 1
x+1 = ±√35
x = -1±√35
The points you seek are
(-1-√35, 0) and (-1+√35, 0), about (-6.916, 0) and (4.916, 0).