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andrew-mc [135]
3 years ago
11

0.20m + 1(50-m) = 0.60(5) 0.80m + 0(50-m) = 0.40(5) please solve for m

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
5 0

Answer:

Step-by-step explanation:

0.20m+50-m=3

0.20m-m=-47

-0.80m=-47

m=58.75

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How many solutions does the following equation have 3(X + 9) = 2x + 54
kotykmax [81]

Answer:

One solution.

Step-by-step explanation:

3(x + 9) = 2x + 54

3x + 27 = 2x + 54

3x + 27 - 2x = 2x + 54 - 2x

x + 27 = 54

x + 27 - 27 = 54 - 27

x = 27

8 0
3 years ago
How do you write this in standard form? (6-7i)/(1-2i)
xenn [34]
\frac { \left( 6-7i \right)  }{ \left( 1-2i \right)  } \cdot 1\\ \\ =\frac { \left( 6-7i \right)  }{ \left( 1-2i \right)  } \cdot \frac { \left( 1+2i \right)  }{ \left( 1+2i \right)  } \\ \\ =\frac { 6+12i-7i-14{ i }^{ 2 } }{ 1+2i-2i-4{ i }^{ 2 } }

\\ \\ =\frac { 6+5i-14\left( -1 \right)  }{ 1-4\left( -1 \right)  } \\ \\ =\frac { 6+5i+14 }{ 1+4 } \\ \\ =\frac { 20+5i }{ 5 } \\ \\ =\frac { 20 }{ 5 } +\frac { 5i }{ 5 } \\ \\ =4+i
8 0
3 years ago
Marci purchased the rug shown. To the nearest hundredth, what area of floor space will the rug cover? (π = 3.14)
quester [9]

Answer:

63.59

Step-by-step explanation:

5 0
3 years ago
Hi I need help on this!!
MA_775_DIABLO [31]

Answer:

VOLUME OF THE CYLINDER=πr²h

=3.14×4²×9

=3.14×16×9

=452.16 mm³

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
3 years ago
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