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babymother [125]
4 years ago
11

What are the missing parts that correctly complete the proof? Given: Point A is on the perpendicular bisector of segment P Q. Pr

ove: Point A is equidistant from the endpoints of segment P Q. Image: A horizontal line segment P Q. A midpoint is drawn on segment P Q labeled as X. A vertical line X A is drawn. A is above the horizontal line. The angle A X Q is labeled a right angle. The line segments P X and Q X are labeled with a single tick mark. A dotted line is used to connect point P with point A. Another dotted line is used to connect point Q with point A. Drag the answers into the boxes to correctly complete the proof. Statement Reason 1. Point A is on the perpendicular bisector of PQ¯¯¯¯¯. Given 2. PX¯¯¯¯¯≅QX¯¯¯¯¯¯ Response area 3. ∠AXP and ∠AXQ are right angles. Response area 4. Response area All right angles are congruent. 5. AX¯¯¯¯¯≅AX¯¯¯¯¯ Reflexive Property of Congruence 6. △AXP≅△AXQ Response area 7. ​ AP¯¯¯¯¯≅AQ¯¯¯¯¯ ​ Response area 8. Point A is equidistant from the endpoints of PQ¯¯¯¯¯. Definition of equidistant Look below me

Mathematics
1 answer:
nikdorinn [45]4 years ago
6 0

Answer:

I just finished the test with an 82. Here are the answers

I hope this helps and brainliest would be greatly appreciated

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Consider the following two ordered bases of R3:
grigory [225]

Answer:

Let A = (a_1, ..., a_n) and B = (b_1, ..., b_n) bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors b_1, ..., b_n at base A.

The, we case correspond to find the coordinates of vectors of C,

\{\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right], \left[\begin{array}{ccc}2\\0\\-1\end{array}\right], \left[\begin{array}{ccc}-3\\1\\2\end{array}\right]   \}

at base B.

1. We need to find a,b,c\in\mathbb{R} such that

\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]

Then we find these values solving the linear system

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

Using rows operation we obtain the echelon form of the matrix

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

now we use backward substitution

c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4

Then the coordinate vector of \left[\begin{array}{ccc}2\\-1\\-1\end{array}\right] is \left[\begin{array}{ccc}4\\2\\1\end{array}\right]

2. We need to find a,b,c\in\mathbb{R} such that

\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]

Then we find these values solving the linear system

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

Using rows operation we obtain the echelon form of the matrix

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

now we use backward substitutionc=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4

Then the coordinate vector of \left[\begin{array}{ccc}2\\0\\-1\end{array}\right] is \left[\begin{array}{ccc}4\\3\\2\end{array}\right]

3. We need to find a,b,c\in\mathbb{R} such that

\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]

Then we find these values solving the linear system

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

Using rows operation we obtain the echelon form of the matrix

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

now we use backward substitutionc=-2\\-b+c=2,\; b=-4\\a-2b+2c=2,\; a=-2

Then the coordinate vector of \left[\begin{array}{ccc}-3\\1\\2\end{array}\right] is \left[\begin{array}{ccc}-2\\-4\\-2\end{array}\right]

Then the change of basis matrix from B to C is

\left[\begin{array}{ccc}4&4&-2\\2&3&-4\\1&2&-2\end{array}\right]

4 0
4 years ago
When sohring the equation, which is the best first step to begin to simplify the equation?
Leno4ka [110]

Answer:

see below

Step-by-step explanation:

-2(x+3)=-10

There are two ways to approach solving this problem

1.  You can distribute -2 to each term in the parentheses

-2x -6 = -10

OR

Divide each side by -2

2. -2(x+3)=-10

  -2/ -2 (x+3) = -10/-2

          x+3 = 5

They will both result in the same answer at the end

7 0
4 years ago
Help me ASAP! ANSWER ALL
Anna [14]

Answer:

Step-by-step explanation:

2. I think C

3.D

4.D

5.B

6. D

7.B

8.A

9.A I think

6 0
3 years ago
X+2y=2 and -x+y=-5. Please help
salantis [7]

Answer:

x=4

y=-1

Step-by-step explanation:

x + 2y = 2

-x + y = -5

3y = -3

y = -1

If you would then plug in -1 (for y) then you would get x=4

For example:

x + 2(-1) = 2

x - 2 = 2

x = 4

OR

-x - 1 = -5

-1 + 5 = x

4 = x

x = 4

5 0
3 years ago
What is the first step in solving this system of equations by Substitution?
Finger [1]

Answer:

you nice

Step-by-step explanation:

hi 3 days

6 0
4 years ago
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