Yes, but on Tuesday she read more because of more time then on Monday
It would be A because they’re both perfect squares
The area of a circular segment is given by
.. A = (1/2)r^2*(θ -sin(θ)) . . . . . . . . . θ in radians
.. = (1/2)*(27.8 in)^2*(5π/6 -sin(5π/6))
.. ≈ 818.4 in^2
Answer:
This is proved by ASA congruent rule.
Step-by-step explanation:
Given KLMN is a parallelogram, and that the bisectors of ∠K and ∠L meet at A. we have to prove that A is equidistant from LM and KN i.e we have to prove that AP=AQ
we know that the diagonals of parallelogram bisect each other therefore the the bisectors of ∠K and ∠L must be the diagonals.
In ΔAPN and ΔAQL
∠PNA=∠ALQ (∵alternate angles)
AN=AL (∵diagonals of parallelogram bisect each other)
∠PAN=∠LAQ (∵vertically opposite angles)
∴ By ASA rule ΔAPN ≅ ΔAQL
Hence, by CPCT i.e Corresponding parts of congruent triangles PA=AQ
Hence, A is equidistant from LM and KN.
Answer:
23 inches
Step-by-step explanation:
If we add 5 inches to the shortest side, all sides will be the same length and the perimeter will be 69 inches. The longest sides have length that is 1/3 that, or ...
... (1/3)·(64 +5 in) = 23 in
_____
You can solve for x, but you obviously don't need to.
The perimeter is ...
... (4x -2) + 2(4x -2 +5) = 64
... 12x +4 = 64
... 12x = 60
... x = 5
... 4x -2 +5 = 4·5 +3 = 23 . . . . the length of the longest side in inches