<span>In this problem, to find the answer we have to setup a series of ratios that relate the scale to real life distance. We know that 1cm = 2.50km, so that ratio would be 1cm/2.5km. For two towns that are 4.75cm apart on the map, we set a ration of 4.75cm/x km, where x is the actual distance. Now we set the ratios equal to each other and solve for x. 1/2.5=4.75/x where x = 4.75*2.5/1 = 11.875 and rounding up we get 11.88 km. The two towns are actually 11.88 km apart from each other.</span>
Option D. D has the matrix of constants [[12], [11], [4]].
Step-by-step explanation:
Step 1:
With the given equations, we can form matrices to represent them.
The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.
Step 2:
The linear system A is represented as
![\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%261%261%5C%5C1%26-11%260%5C%5C1%26-1%264%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\z\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D26%5C%5C17%5C%5C23%5Cend%7Barray%7D%5Cright%5D)
.
Step 3:
The linear system B is represented as
![\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%261%261%5C%5C1%26-11%260%5C%5C1%26-1%2612%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D23%5C%5C17%5C%5C26%5Cend%7Barray%7D%5Cright%5D)
.
Step 4:
The linear system C is represented as
![\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%5C%5C1%26-1%260%5C%5C1%26-1%261%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C11%5C%5C12%5Cend%7Barray%7D%5Cright%5D)
.
Step 5:
The linear system D is represented as
![= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%5C%5C11%5C%5C4%5Cend%7Barray%7D%5Cright%5D)
.
Step 6:
Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.
Pythagorean Theorem is
a ²+ b ² = c ²
a and b are the shorter sides, while c is the hypotenuse, or longest side. Let’s say the side we are looking for is “b”.
Let’s first rearrange the equation, then sub in.
a ²+ b ² = c ²
b ² = c ²- a ²
b = √ (c ²- a ²)
b = √(17 ²- 15 ²)
b = √(289-225)
b = √ 64
b = 8
The dimensions of the rectangle are 150 m and 75 m respectively.
Let x be the length of the rectangular area and y be the width of the rectangular area.
Its area A = xy
Given that its area is 11,000 m², xy = 11000 m² (1)
Also, since the bounding side of the building is 150 m long and the fencing is 300 m, the perimeter of the rectangular area is P = 150 m + 300 m = 450 m.
Also, the perimeter of the rectangular area is P = 2(x + y)
Since P = 450 m
2(x + y) = 450 m
x + y = 225 m (2)
Since the bounding side of the building represents one dimension of the rectangle, x = 150 m.
Substituting x into (2), we get the other dimension
x + y = 225
150 + y = 225
y = 225 m - 150 m
y = 75 m
So, the dimensions of the rectangle are 150 m and 75 m respectively.
Learn more about dimensions of a rectangular area here:
brainly.com/question/16818606
The rate of how many projects Taylor can complete can be modeled at x/6 and Brianne can be modelled as x/8. So, by combining the functions and setting equal to 1, you can solve.
x/6 + x/8 = 1
You have to give common demoninators
4x/24 + 3x/24 = 1
Then add together and cross multiply. Giving you an answer of 24/7 or roughly 3.43 hours when rounded to the nearest hundredth.
