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Marianna [84]
2 years ago
7

The right isosceles triangle PQR has vertices P(3, 3), Q(3, 1) and R(x, y) and is rotated 90 degrees counterclockwise about the

origin. Find the missing vertex of the triangle. Then graph the triangle and its image.
Mathematics
1 answer:
Jlenok [28]2 years ago
7 0
For the answer to the question above, since the triangle is isosceles, the two legs have equal length. The coordinates of two vertices are given
P(3,3)
Q(3,1)
Assuming that PQ and QR are the legs of equal length, the distance between Q and R must be the same as the distance between P and Q
d = √[(3-3)² + (3-1)²
d = 2
Therefore, the coordinates of P is
(5,1)
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Ainat [17]
Answer choice 3 is correct
The answer is 3x+18=147°, because a central angle is equal to the amount of degrees of its arc.
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2 years ago
4) What is the present value of $4000 received at the
Bingel [31]

Answer:

$11040

Step-by-step explanation:

first of all the question says that $4000 were earned in a year and asks for what the new vale would be after the next 3 years with a discount rate of 8%.

If 1 year=$4000,then 3 years=$12000

100%-8%=92% (this happens because there is still a remaining amount that still has a cost to it),so 12000*92%=$11040

8 0
3 years ago
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) Find the coordinates of the point A' which is the symmetric point
soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point  A(3, 2) with respect to the line 2x + y - 12 = 0.​

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

C( \dfrac{x+3}{2}, \dfrac{y+2}{2})

Also, product of slope will be -1 .( Since, they are parallel )

\dfrac{y-2}{x-3} \times -2  = -1\\\\2y - 4 = x - 3\\\\2y - x = 1

x = 2y - 1

So, C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})

Also, C satisfy given line :

2\times ( \dfrac{2y + 2}{2})  + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}

Also,

x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}

Therefore, the symmetric points isA'(\dfrac{31}{5}, \dfrac{18}{5}) .

7 0
2 years ago
How do you rotate the coordinate (9,-6) 270 degrees counterclockwise?
GalinKa [24]
Since this point is located in the fourth quadrant you have to rotate it 270 degrees counterclockwise which would end up being in third quadrant. which means the new point would be (-6,-9)
6 0
3 years ago
Reflect triangle ABC over the x-axis. Then translate the new triangle 5 units to the left. What are the coordinates of the verti
makvit [3.9K]

Hi there!

The answer to your question is B my explanation is below

The coordinates of the vertices of the final image will be:

A''(-6, 8), B''(-7, -2), and C''(0, -2)

Hence, option (B) is true.

Step-by-step explanation:

Given the triangle with the vertices

A(-1, -8)B(-2, 2)C(5, 2)We know that when we reflect a point across the x-axis, the x-coordinate remains the same, but the y-coordinate reverses its sign.

Thus,

The rule of reflection across the x-axis:

(x, y) → (x, -y)

so

A(-1, -8) → A'(-1, -(-8)) = A'(-1, 8)

B(-2, 2) → B'(-2, -2) = B'(-2, -2)

C(5, 2) → C'(5, -2) = C'(5, -2)

Then, translate the new triangle 5 units to the left

Rule of translating 5 units to the left

(x, y) → (x-5, y)

so

A'(-1, 8) → A''(-1-5, 8) = A''(-6, 8)

B'(-2, -2) → B''(-2-5, -2) = B''(-7, -2)

C'(5, -2) → C''(5-5, -2) = C''(0, -2)

Therefore, the coordinates of the vertices of the final image will be:

A''(-6, 8), B''(-7, -2), and C''(0, -2)

Hence, option (B) is true.

8 0
3 years ago
Read 2 more answers
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