Let O be the point of intersection of diagonals.
Consider triangles FOE and EOD:
FO = OD, ∠FOE = ∠EOD and OE is a common side ⇒
triangles FOE and EOD are congruent.
In congruent triangles all corresponding sides and are congruent ⇒
EF = DE = 6
i. Let t be the line tangent at point J. We know that a tangent line at a point on a circle, is perpendicular to the diameter comprising that certain point. So t is perpendicular to JL
let l be the tangent line through L. Then l is perpendicular to JL ii. So t and l are 2 different lines, both perpendicular to line JL.
2 lines perpendicular to a third line, are parallel to each other, so the tangents t and l are parallel to each other.
Remark. Draw a picture to check the
Answer:
Whats the question?
Step-by-step explanation:
5. A. (4, -2)
6. C. (x, y) — (x, -y + 5)
Step-by-step explanation:
5. For the formula y = x, the x and y coordinates get swapped.
M = (-2, 4) — M’ = (4, -2)
6. If the coordinates get reflected across the x-axis, the y coordinates become negative.
(x, y) — (x, -y)
Now that the coordinates are reflected, you go 5 units up (+ 5) to get to the reflection of the coordinates if it was 5 units down before it reflected across the x-axis (- 5).
Ex. 1, 6 gets reflected across the x-axis and moved 5 units up. It’s reflection would be equivalent to (1, -1) because it moved 5 units down (1, 1) then reflected across the x-axis (1, -1).
(x, y - 5) reflected across the x-axis is equivalent to (x, -y + 5)
9514 1404 393
Answer:
see attached
Step-by-step explanation:
The filling in of the formula for the n-th term is pretty straightforward. The attachment shows how simple it is.
The 7th term is found by evaluating the expression for n=7.
a₇ = 192