Answer:
Greatest Common Factor is 
Step-by-step explanation:
Greatest Common Factor is the highest factor that will divide all the numbers. To find the Greatest common factor, first we need to find out the numbers that all the values have in common.
Factors of 4k=1, 2, 4 and k
Factors of 18k=1, 2 , 3, 6, 9, 18 and k
Factors of 12=1, 2, 3, 4, 6 and 12
Common factors are = 1 and 2.
Therefore,
Greatest Common Factor = 
=
The ratio of Isabella’s money to Shane’s money is 3:11
Isabella has $33.
We can make a proportion to solve for how much money Shane has.
A proportion is two ratios that are set equal to each other.
Let’s call Shane’s money ‘S’
We get this proportion: 3/11 = 33/x
If we cross multiply we get:
(33) * (11) = (3) * (x)
Simplifying it, we get:
363 = (3) * (x)
Divide both sides by 3, we get:
x = 121
However, the question asks how much money they have together.
Isabella + Shane = Total
33 + 121 = $154
<span>They have $154 together.</span>
Answer:
15
Step-by-step explanation:
The equation is just a plug in and solve. you have your given variables and all you have to do it plug them into the matching spot and solve
R=8 P=7
8+7=15
The difference is 99 percent off and
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 .
