Answer:
iv ; xy + x/y = 3
Step-by-step explanation:
A non linear equation is one in which there is no product of two different variables or the same variable
What this mean is that we have the highest polynomial power as 1 or we simply have polynomials of degree 1
Looking at the given options, we can see the product xy
What this mean is that the polynomial is of degree 2 and that makes the equation a non linear as we have a polynomial of more than 1 degree
{{{ THE BOLDED CHARACTERS SHOULD BE SMALL. }}}
SEQUENCE: 6, 18, 54, 162
18/6 = 3
54/18 = 3
162/54 = 3
then, r (common ratio) = 3
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RECURSIVE RULE: r = 3
an = a(n - 1) × r [formula]
ANSWER: an = a(n - 1) × 3
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ITERIATIVE RULE: r = 3, a1 = 6
an = a1 × r^(n - 1) [formula] [ ^(n-1) is an exponent]
ANSWER: an = 6 × 3^(n - 1)
Dividing the number of tires that should be installed per day which is 400 by the number of working hours which is 8 will give us 50 tires per hour. Assuming that the same mistake will take toll on the workers such that they will have 1 tire mistakenly installed in an hour, they will have 8 erroneous tires in a day. Multiplying this by 5 to make the answer per week will give 40. Out of the 400 x 5 = 2000 tires. The answer would be 2000 - 40 which is equal to 1940. The assumption must be valid.
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 .
