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3 years ago

Two quantities a and b have a proportional relationship where a is -4 and b -3 create an equation to represent this relationship

Mathematics

1 answer:

sladkih [1.3K]3 years ago

6 0

Answer:

a = (4/3)b

Step-by-step explanation:

The governing equation here has the form y = kx, where k is the constant of proportionality.

Let's rewrite that as a = kb, and then substitute -4 for a and -3 for b:

-4 = k(-3), or k = 4/3

Then a = (4/3)b

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If 5x+6=10, what is the value of 10x+3? A) 4 B) 9 C) 11 D) 20

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3 years ago

Can you tell me how to calculate R for below Matrix ?

scoray [572]

Answer:

The rank of the matrix is 3.

Step-by-step explanation:

Consider the prided information.

$x=\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}=\begin{bmatrix}9&2&6&5&8\\ 12&8&6&4&10\\ 3&4&0&2&1\end{bmatrix}$

Reduce the matrix in row echelon form as shown:

$R_1\:\leftrightarrow \:R_2\begin{bmatrix}12&8&6&4&10\\ 9&2&6&5&8\\ 3&4&0&2&1\end{bmatrix}\\$

$R_2\:\leftarrow \:R_2-\frac{3}{4}\cdot \:R_1\ and\ R_3\:\leftarrow \:R_3-\frac{1}{4}\cdot \:R_1\\\\\begin{bmatrix}12&8&6&4&10\\ 0&-4&\frac{3}{2}&2&\frac{1}{2}\\ 0&2&-\frac{3}{2}&1&-\frac{3}{2}\end{bmatrix}$

$R_3\:\leftarrow \:R_3+\frac{1}{2}\cdot \:R_2\\\begin{bmatrix}12&8&6&4&10\\ 0&-4&\frac{3}{2}&2&\frac{1}{2}\\ 0&0&-\frac{3}{4}&2&-\frac{5}{4}\end{bmatrix}$

$R_2\:\leftarrow \:R_2-\frac{3}{2}\cdot \:R_3\ and\ R_1\:\leftarrow \:R_1-6\cdot \:R_3\\\begin{bmatrix}12&8&0&20&0\\ 0&-4&0&6&-2\\ 0&0&1&-\frac{8}{3}&\frac{5}{3}\end{bmatrix}\\R_2\:\leftarrow \:-\frac{1}{4}\cdot \:R_2\ , \ R_1\:\leftarrow \:R_1-8\cdot \:R_2\ and\ \:R_1\:\leftarrow \frac{1}{12}\cdot \:R_1\\\begin{bmatrix}1&0&0&\frac{8}{3}&-\frac{1}{3}\\ 0&1&0&-\frac{3}{2}&\frac{1}{2}\\ 0&0&1&-\frac{8}{3}&\frac{5}{3}\end{bmatrix}$

The rank of a matrix is the number of non zeros rows.

Thus, the rank of the matrix is 3.

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3 years ago