Find the value that makes the pair of ratios equivalent. 12:42,2:x my answers are 8, 7, 6 and 5.

Mathematics

1 answer:

Marianna [84]3 years ago

4 0

The answer is option 6! hope it helps

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Simplify the expression

damaskus [11]

First, simplify the denominator into 1+3i. Then multiply both the numerator and denominator by 1-3i to rationalize the denominator. Then, the numerator becomes i+3, and the denominator becomes 10. So, the first answer is correct.

3 0

3 years ago

How do you solve? <br>2(1+4n)+10(n-7)=2n-n

stiv31 [10]

<span>2(1+4n)+10(n-7)=2n-n
distribute
2+8n+10n-70=2n-n
combine like terms
18n-68=n
move n to other side
17n=68
divide
n=4</span>

4 0

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:

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

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

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

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