There is no way that he can run 151515km/h, but if he really can, then :
1 hour = 60 minutes
(151515km/h)/60minutes = 2525,25km/minute
Answer:
To construct the Square using a compass and straight edge, you have written the steps not in proper order. I am rearranging the steps for you
1. Construct horizontal H J¯¯¯¯¯
2. Construct a circle with point H as the center and a circle with point J as a center with each circle having radius HJ .
3.Label the point of intersection of the two circles above HJ¯¯¯¯¯ , point K, and the point of intersection of the two circles below HJ¯¯¯¯¯ , point L.
4.Construct KL¯¯¯¯¯ , the perpendicular bisector of HJ¯¯¯¯¯ , intersecting HJ¯¯¯¯¯ at point M.
5.Construct a circle with point M as the center with radius MJ .
6.Label the point of intersection of circle M and KL¯¯¯¯¯ closest to point K, point N, and the point of intersection of circle M and KL¯¯¯¯¯ closest to point L, point O.
7.Construct HN¯¯¯¯¯¯ , NJ¯¯¯¯¯ , JO¯¯¯¯¯ , and OH¯¯¯¯¯¯ to complete square HNJO .
Answer:
Step-by-step explanation:
a taingle adds up to 180 degrees if that helps
The correct answer is: "(-1, -2)" .
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Note: By examining the point of intersection of the lines of the 2 graphs show in the "image attached", we can see that the point of intersection is at: "(-1, 2"); that is: "x = -1, y = 2" .
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Answer:
(A) Set A is linearly independent and spans 
. Set is a basis for .
Step-by-Step Explanation
<u>Definition (Linear Independence)</u>
A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.
<u>Definition (Span of a Set of Vectors)</u>
The Span of a set of vectors is the set of all linear combinations of the vectors.
<u>Definition (A Basis of a Subspace).</u>
A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.
Given the set of vectors ![A= \left(\begin{array}{[c][c][c][c]}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\end{array} \right)](https://tex.z-dn.net/?f=A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7B%5Bc%5D%5Bc%5D%5Bc%5D%5Bc%5D%7D1%20%26%200%20%26%200%20%26%200%5C%5C%200%20%26%201%20%26%200%20%26%201%5C%5C%200%20%26%200%20%26%201%20%26%201%5Cend%7Barray%7D%20%5Cright%29%20)
, we are to decide which of the given statements is true:
In Matrix ![A= \left(\begin{array}{[c][c][c][c]}(1) & 0 & 0 & 0\\ 0 & (1) & 0 & 1\\ 0 & 0 & (1) & 1\end{array} \right)](https://tex.z-dn.net/?f=A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7B%5Bc%5D%5Bc%5D%5Bc%5D%5Bc%5D%7D%281%29%20%26%200%20%26%200%20%26%200%5C%5C%200%20%26%20%281%29%20%26%200%20%26%201%5C%5C%200%20%26%200%20%26%20%281%29%20%26%201%5Cend%7Barray%7D%20%5Cright%29%20)
, the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column. has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans .
Therefore Set A is linearly independent and spans . Thus it is basis for .
