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Bezzdna [24]
3 years ago
7

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

coordinates B’ and C’ of it’s image are (–2, 2) and (1, 4). What is the coordinate of vertex A. (EXPLAIN WORK)

Mathematics
1 answer:
Lemur [1.5K]3 years ago
8 0

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates (x_A,y_A)

Vectors AB and AB' are perpendicular, then

\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0

Vectors AC and AC' are perpendicular, then

\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0

Now, solve the system of two equations:

\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.

Subtract these two equations:

5x_A-y_A-8=0\Rightarrow y_A=5x_A-8

Substitute it into the first equation:

x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2

Then

y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

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kifflom [539]

Answer:

Ammm, I think the answer you have is wrong, but what I really think is that the equations you put in are not the right ones because the answer for these is "ugly" (decimals etc...). Want to check your equations?

Step-by-step explanation:

Here is a step by step explanation to your equations (I assumed +3z for the second but you can change it). You can edit it for your equations if you want to.

https://simplisico.com/share/q/QiWKOyN5uYgkSG4iBXOmdG0i

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2 years ago
What is the solution to the given inequality?
SashulF [63]

1/2 - 1/4x ≥ -1/4

Combine 1/4x to get x/4:

1/2 - x/4 ≥ -1/4

Subtract 1/2 from both sides:

-x/4 ≥ -1/4 - 1/2

Simplify:

-x/4 ≥ -3/4

Multiply both sides by 4:

-x ≥ -3/4 * 4

-x ≥ -3

Multiply each side by -1 ( and since you are multiplying by -1, you also need to reverse the inequality sign)

Answer: x ≤ 3

5 0
3 years ago
Which figure has one line of symmetry but no rotational symmetry?
Slav-nsk [51]
C because u can split it only once vertically but it’s not a rotationally symmetrical shape
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3 years ago
The number of square feet per house is normally distributed with a population standard deviation of 154 square feet and an unkno
11111nata11111 [884]

Answer: (1500.66,1599.34)

Step-by-step explanation:

Confidence interval for population mean:

\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}

,where \overline{x} = Sample mean , n= Sample size, z* = critical two tailed z-value , \sigma = population standard deviation.

As per given , we have

n= 16

\sigma = 154 square feet

\overline{x}= 1550 square feet

\alpha= 1-0.80 = 0.20

Critical z-value = z_{\alpha/2}=z_{0.2/2}=z_{0.1}=1.2815

Confidence interval for population mean:

1550\pm (1.2815)\dfrac{154}{\sqrt{16}}\\\\ = 1550\pm (1.2815)\dfrac{154}{4}\\\\= 1550\pm 49.33775\\\\ =(1550-49.33775,\ 1550+49.33775)\\\\\approx (1500.66,\ 1599.34)

Required confidence interval:  (1500.66,1599.34)

6 0
2 years ago
A circular swimming pool has a radius of 15 ft. There is a path all around that pool that is three feet wide.
Romashka-Z-Leto [24]

Answer:

Circumference(C) of the circle is given by:

C = 2 \pi r

where, r is the radius of the circle.

As per the statement:

A circular swimming pool has a radius of 15 ft.

⇒radius of the pool(r) = 15 ft

There is a path all around that pool that is three feet wide.

⇒radius of the outer edge of the path around the pool(r') = 15 +3 = 18 ft

Substitute these in the formula we have;

C = 2 \pi r'

⇒C = 2 \cdot 3.14 \cdot 18 = 113.04 ft

Therefore, the circumference of the outer edge of the path around the pool is, 113.04 ft

6 0
3 years ago
Read 2 more answers
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