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adelina 88 [10]
4 years ago
8

Reflect triangle A in the line y = 1.

Mathematics
1 answer:
Over [174]4 years ago
3 0

Answer:

Step-by-step explanation:

In the figure attached,

Vertices of the triangle ABC are A(-4, 4), B(-1, 4) and (-1, 2)

When we reflect this figure across a line y = 1

Vertices of the image triangle A'B'C' will be

Coordinates of A'

A(-4, 4) → A'[(1 + 5), 4]

            → A'(6, 4)

Coordinates of B'

B(-1, 4) → B'(-1 + 4, 4)

           → B'(3, 4)

Coordinates of C',

C(-1, 2) → C'[(-1 + 4), 2]

           → C'(3, 2)

Graph these points and join them to form the image triangle A'B'C'.

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Step-by-step explanation:

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Is the set of numbers one 1,5,25 and 125 a geometric sequence
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Find the general equation.
tn = a*5^(n - 1)

Example
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3 years ago
What is the distance from P to Q ?A. 7 unitsB. 5 unitsC. 1 unitD. 25 unitshow do I find the distance from P to Q?
Alex17521 [72]

to find the distance between 2 points we should apply the formula

d=\sqrt[]{(x_2-x_1)^2+(y_2-_{}y_1)^2_{}}

call point q as point 1 for reference in the formula and p as point 2

replace the coordinates in the formula

d=\sqrt[]{(3-(-1))^2+(-4-(-1))^2}

simplify the equation

\begin{gathered} d=\sqrt[]{(3+1)^2+(-4+1)^2} \\ d=\sqrt[]{4^2+(-3)^2} \\ d=\sqrt[]{16+9} \\ d=\sqrt[]{25} \\ d=5 \end{gathered}

the distance between the 2 points is 5 units

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