In each table, x increases by 1. We start with x = 0 and stop with x = 3. So we will focus on the y columns of each table as those are different.
Let's move from left to right along the four tables.
For the first table, we go from y = 1 to y = 2. That's an increase of 1
Sticking with the first table, we go from y = 2 to y = 4. The increase is now 2
Since the increase is not the same, this means the table is not linear. The y increase must be constant. We can rule out choice A
Choice B can be ruled out as well. Why? Because...
the jump from y = 0 to y = 1 is +1
the jump from y = 1 to y = 3 is +2
The same problem comes up as it did with choice A
Choice C has the same problem, but the increase turns into a decrease half the time. We go from y = 0 to y = 1, then we go back to y = 0 so the "increase" is really a decrease. We can think of it as a negative increase. Regardless, this allows us to rule out choice C
Only choice D is the answer. Each time x goes up by 1, y goes up by 2. Therefore the slope is 2/1 = 2
You are given Kayla's walking distance along the edge of the river 100 ft and marks point C, then she walks 100 ft further and marks point D and she turns 90° and walks until her location, point A, and point C are collinear. She marks point E at this location. ABC and EDC are congruent. thats just what i think it is tho
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.
You can square the whole problem to cancel out the square root. Might make things easier.
The area of the sector is 472.6 ft²
<h3>Area of a Sector</h3>
For finding the area of the sector, you need apply the formula:
, where:
r= radius
α= central angle
The question gives:
r= radius= 19 ft
α= central angle = 150°
Thus, the area of the sector will be:
Read more about the area of a sector here:
brainly.com/question/22972014
#SPJ1
