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Komok [63]
3 years ago
8

What is the fraction for What is the fraction for 0.21

Mathematics
1 answer:
gavmur [86]3 years ago
3 0
The fraction for 0.21 would be 21/100
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Is the discriminant of f positive , zero , or negative ​
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Answer:

Negative because there's no x-intercepts!

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8 0
3 years ago
Determine whether the set of vectors is a basis for ℛ3. Given the set of vectors , decide which of the following statements is t
schepotkina [342]

Answer:

(A) Set A is linearly independent and spans R^3. Set is a basis for R^3.

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) , 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) , 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. R^3 has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans R^3.

Therefore Set A is linearly independent and spans R^3. Thus it is basis for R^3.

8 0
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Please help me with this problem
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Sorry if this is messy but I hope it helps you

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Bread:whole wheat,Meat: turkey and Cheese: cheddar

Step-by-step explanation:

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(-2,-3)(0,-5)(3,-8)

Step-by-step explanation:

6 0
3 years ago
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