Answer:
30√3 ≈ 51.96 miles
Step-by-step explanation:
The distance between the two ships can be found using the Law of Cosines, or using your knowledge of the side relationships in special triangles.
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Each ship is traveling at 10 mph, so after 3 hours will have traveled 30 miles.
The triangle OS1S2 formed by the harbor and the two ship locations is an isosceles triangle with base angles of 30°. Each half of OS1S2 is a 30-60-90 triangle whose longer leg is √3 times half the hypotenuse. The sum of those two "longer legs" is the distance between the ships.
The distance between ships is 2×15√3 = 30√3 ≈ 51.96 miles.
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<em>Additional comment</em>
If you prefer to use the Law of Cosines, you are looking for the length of the side opposite the 120° angle in a triangle with sides of 30 miles.
c² = 30² +30² -2·30·30·cos(120°) = 30²(2-2·(-0.5)) = 3·30²
c = 30√3 . . . . . take the square root (miles)
F = 9. pretty easy solution, just whatever you do to one side u do to the other.
Given parameters:
Midpoint of AB = M(3, -1)
Coordinates of A = (5,1)
Unknown:
Coordinates of B = ?
Solution:
To find the mid point of any line, we use the expression below;
and 
where 
and 
= coordinates of the mid points = 3 and -1
x₁ = 5 and y₁ = 1
x₂ = ? and y₂ = ?
Now let us input the variables and solve,
3 = 
and -1 = 
5 + x₂ = 6 -2 = 1 + y₂
x₂ = 1 y₂ = -2 -1 = -3
The coordinates of B = 1, -3
Find a basis for the column space and rank of the matrix ((-2,-2,-4,1),(7,-3,14,-6),(2,-2,4,-2)(2,-6,4,-3))
Novosadov [1.4K]
Answer:
- B=\{\left[\begin{array}{c}-2\\-2\\-4\\1\end{array}\right], \left[\begin{array}{c}7\\-3\\14&-6\end{array}\right], \left[\begin{array}{c}2\\-2\\4\\-2\end{array}\right] \}[/tex] is a basis for the column space of A.
- The rank of A is 3.
Step-by-step explanation:
Remember, the column space of A is the generating subspace by the columns of A and if R is a echelon form of the matrix A then the column vectors of A, corresponding to the columns of R with pivots, form a basis for the space column. The rank of the matrix is the number of pivots in one of its echelon forms.
Let ![A=\left[\begin{array}{cccc}-2&7&2&2\\-2&-3&-2&-6\\-4&14&4&4\\1&-6&-2&-3\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-2%267%262%262%5C%5C-2%26-3%26-2%26-6%5C%5C-4%2614%264%264%5C%5C1%26-6%26-2%26-3%5Cend%7Barray%7D%5Cright%5D)
the matrix of the problem.
Using row operations we obtain a echelon form of the matrix A, that is
![R=\left[\begin{array}{cccc}1&-6&-2&-3\\0&9&2&0\\0&0&-8&-21\\0&0&0&0\end{array}\right]](https://tex.z-dn.net/?f=R%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-6%26-2%26-3%5C%5C0%269%262%260%5C%5C0%260%26-8%26-21%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%5D)
Since columns 1,2 and 3 of R have pivots, then a basis for the column space of A is
![B=\{\left[\begin{array}{c}-2\\-2\\-4\\1\end{array}\right], \left[\begin{array}{c}7\\-3\\14&-6\end{array}\right], \left[\begin{array}{c}2\\-2\\4\\-2\end{array}\right] \}](https://tex.z-dn.net/?f=B%3D%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-2%5C%5C-2%5C%5C-4%5C%5C1%5Cend%7Barray%7D%5Cright%5D%2C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D7%5C%5C-3%5C%5C14%26-6%5Cend%7Barray%7D%5Cright%5D%2C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-2%5C%5C4%5C%5C-2%5Cend%7Barray%7D%5Cright%5D%20%5C%7D)
.
And the rank of A is 3 because are three pivots in R.
