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worty [1.4K]
3 years ago
13

How do i prove that measurement of angle abc is equal to measurement of def

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
6 0
The lines are parallel to each other therefore so are the angles.
You might be interested in
A printer prints 75 pages in 5 minutes. At the same rate, how many pages does the printer print in 7 minutes? Solve and show you
Bingel [31]

Answer:

105

Step-by-step explanation:

first, the printer print 15  (75/5) pages per minutes,

next, 7×15=105

8 0
3 years ago
What are the coordinates of P' when you translate P 4 units to the right and 3 unit down?
OLga [1]

Answer:

(3,2)

Step-by-step explanation:

The coordinates of the point P on the given graph is (-1,5).

So, when we translate point P on the graph by 4 units to the right its x-coordinate will change to (- 1 + 4) = 3.

Again, when we translate point P on the graph by 3 units down then its y-coordinate will change to (5 - 3) = 2.

Therefore, the new coordinates of the point P will be (3,2) (Answer)

4 0
3 years ago
5x+3&lt;7+7x<br> Please explain in clear details
trasher [3.6K]

Answer:

5x just means 5 times so, 5 times a number(or variable)plus three is less than 7 times a number variable)plus 7

4 0
3 years ago
Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you
adell [148]

Answer:

1) The solution of the system is

\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right

2) The solution of the system is

\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right

Step-by-step explanation:

1) To solve the system of equations

\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 2: Swap rows 1 and 2

\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 3:  \left(R_1=\frac{R_1}{6}\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 4: \left(R_3=R_3-R_1\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]

Step 5: \left(R_2=\frac{R_2}{3}\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]

Step 6: \left(R_3=R_3+R_2\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right]

Step 7: \left(R_3=\left(\frac{6}{37}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 8: \left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 9: \left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 10: Rewrite the system using the row reduced matrix:

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right

2) To solve the system of equations

\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right

using the row reduction method you must:

Step 1:

\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 2: \left(R_1=\frac{R_1}{4}\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 3: \left(R_2=R_2-\left(2\right)R_1\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 4: \left(R_3=R_3-R_1\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 5: \left(R_2=\left(2\right)R_2\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 6: \left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 7: \left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right]

Step 8: \left(R_3=\left(- \frac{2}{117}\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 9: \left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 10: \left(R_2=R_2-\left(15\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 11:

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right

8 0
3 years ago
In math class, you have the following scores: 77, 82, and 73. If the
ziro4ka [17]

Answer: 88

Step-by-step explanation: 77+82+73= 232. So you need an 80 for average. So what you have to do is 80x4 and that equals to 320. last step is to subtract 320-232 and you get 88. 77+82+73+88=320.

6 0
3 years ago
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