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vitfil [10]
3 years ago
8

Reflect the given preimage over y=−1 followed by y=−7. Find the new coordinates. What one transformation is this double reflecti

on the same as? (Note: when you are reflecting over a y= line, the x-values of the preimage will remain the same and you will be changing the y-values)
Mathematics
1 answer:
fenix001 [56]3 years ago
5 0

Answer:

Reflecting over y = -1 line:

A'(8, -10)

B'(10, -8)

C'(2, -4)

Reflecting over y = -7 line:

A''(8, -4)

B'(10, -6)

C''(2, -10)

Step-by-step explanation:

Reflect the given preimage over y=−1 followed

by y=−7. Find the new coordinates. What one transformation is this double reflection the same as? (Note: when you are reflecting over a y= line, the x-values of the preimage will remain the same and you will be changing the y-values)

The coordinates of the preimage are:

A(8,8)

B(10,6)

C(2,2)

Answer: Reflecting over y = -1:

If a point is reflected over a y line, the x values remain the same while the y values change.

For point A(8, 8): The y distance between the y = - 1 line and point A is 9 units. (8- (-1)). If point A is reflected, the y value would be 9 units below the y = -1 line, i.e the new y coordinate would be at -10 (-1-9)

The new coordinate is at A'(8, -10)

For point B(10, 6): The y distance between the y = - 1 line and point B is 7 units. (6- (-1)). If point B is reflected, the y value would be 7 units below the y = -1 line, i.e the new y coordinate would be at -8 (-1-7)

The new coordinate is at B'(10, -8)

For point C(2, 2): The y distance between the y = - 1 line and point C is 3 units. (2- (-1)). If point C is reflected, the y value would be 3 units below the y = -1 line, i.e the new y coordinate would be at -4 (-1-3)

The new coordinate is at C'(2, -4)

Reflecting over y = -7 line:

For point A'(8, -10): The y distance between the y = - 7 line and point A' is 3 units. (-7- (-10)). If point A' is reflected, the y value would be 3 units above the y = -7 line, i.e the new y coordinate would be at -4 (-7+3)

The new coordinate is at A''(8, -4)

For point B'(10, -8): The y distance between the y = - 7 line and point B' is 1 units. (-7- (-8)). If point B' is reflected, the y value would be 1 units above the y = -7 line, i.e the new y coordinate would be at -6 (-7 + 1)

The new coordinate is at B'(10, -6)

For point C'(2, -4): The y distance between the y = - 7 line and point C' is 3 units. (-4- (-7)). If point C' is reflected, the y value would be 3 units below the y = -7 line, i.e the new y coordinate would be at -10 (-7-3)

The new coordinate is at C''(2, -10)

We can also see that h= −7−(−1)=−6. We know that two reflections is the same as a translation of 2h  units down. So 2(−6) is a translation of −12 units down.

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First, use the distributive property and split 2 off from 1/2 in the hours category.

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Please please please help me with this answer
Rasek [7]

x = number of packages he bought

y = number of popsicles in each package

Here are the two equations:

xy = 54

x = 3 + y

Using the substitution method, you can solve them to find x and y.

(3 + y)y = 54,

3y + y^2 = 54

y^2 + 3y - 54 = 0

Solve the quadriatic equation:

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y = -9, 6.

Plug y into: xy = 54

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xz_007 [3.2K]

Answer:

397.7 m²

Step-by-step Explanation:

Step 1: find m < W

W = 180 - (33+113) (sum of ∆)

W = 34°

Step 2: find side UV using the law of sines

\frac{UV}{sin(W)} = \frac{VW}{sin(U)}

\frac{UV}{sin(34)} = \frac{29}{sin(33)}

Multiply both sides by sin(34)

\frac{UV}{sin(34)}*sin(34) = \frac{29}{sin(33)}*sin(34)

UV = \frac{29*sin(34)}{sin(33)}

UV = 29.8 m (approximated)

Step 3: find the area using the formula, ½*UV*VW*sin(V)

area = ½*29.8*29*sin(113)

Area = 397.7 m² (rounded to the nearest tenth.

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