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2 years ago

10

2 ratios equivalent to 3:12

2 answers:

Hatshy [7]2 years ago

6 0

Answer: 1 : 4 and 2 : 8

Step-by-step explanation: To find 2 different ratios that are equal to the ratio 3:12, we first rewrite the ratio 3:12 as the fraction 3/12. Notice that we can find 1 of our equal ratios by simply writing this fraction in lowest terms. if we divide both the numerator and denominator of 3/12 by their greatest common factor of 3, we get the equal ratio 1/4.

Now we can use 1/4 to find another equal ratio. If we multiply both the numerator and denominator of 1/4 by 2, we get the equal ratio 2/8.

Finally, remember to write these ratios using the same form as the original problem. 1 : 4 and 2 : 8.

Alborosie2 years ago

3 0

Answer:

6:24

12:48

Step-by-step explanation:

3 : 12

x2

6:24

x2

12:48

:)

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Find all x in set of real numbers R Superscript 4 that are mapped into the zero vector by the transformation Bold x maps to Uppe

sukhopar [10]

Answer:

$x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]$

Step-by-step explanation:

According to the given situation, The computation of all x in a set of a real number is shown below:

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$\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]$

Now the augmented matrix is

$\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]$

After this, we decrease this to reduce the formation of the row echelon

$R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]$

$R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]$

$R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]$

$R_2 = \frac{R_2}{4}, R_3 = \frac{R_3}{-4} \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]$

$R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]$

$R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]$

$= x_1 - 4x_3 = 0\\\\x_1 = 4x_3\\\\x_2 - 3x_3 = 0\\\\ x_2 = 3x_3\\\\x_4 = 0$

$x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]$

By applying the above matrix, we can easily reach an answer

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