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SCORPION-xisa [38]
2 years ago
4

Points S, U, and T are the midpoints of the sides of TrianglePQR. Triangle S U T is inside of triangle P Q R. Points S, U, and T

are the midpoints of triangle P Q R. Which statements are correct? Check all that apply.
Mathematics
2 answers:
Rzqust [24]2 years ago
3 0

Answer:

1/2 QP = UT

SU ║ RP

Step-by-step explanation:

If UT is parallel to QP, then it is half the length of QP. ΔTUR ~ ΔPQR with a scale factor of 1/2.

It T is the midpoint of PR, then SU is parallel to RP and half its length. Again, ΔQSU ~ ΔQPR with a scale factor of 1/2.

_____

There is no diagram accompanying this problem statement. If the above assumptions are incorrect, then the conclusions are incorrect.

ra1l [238]2 years ago
3 0

Answer:

A and D

Step-by-step explanation:

it just is

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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:

First we have to determine the \bar x so that A \bar x = 0

\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

5 0
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