6/2(1+2)+1-1+1-1+1-1
2 answers:
You might be interested in
Answer:
The correct option is A.
Step-by-step explanation:
It is given that the QRST is a parallelogram and the diagonals QS and RT are congruent.
According to the property of parallelogram if the length of diagonals are same, then the parallelogram is rectangle. We can also prove it.
Let QRST be an parallelogram and QS=RT.
In triangles QRT and RQS.
(Diagonals)
(Common side)
(Opposite sides of the parallelogram are same)
So by SSS rule of congruence, triangle QRT and triangle RQS are congruent.
By CPCT,
Since the sum of two consecutive angles of a parallelogram is 180 degree.
Therefore the measure of angle R and angle Q are 90 degree.
Similarly in triangle QST and RTS,
(Diagonals)
(Common side)
(Opposite sides of the parallelogram are same)
So by SSS rule of congruence, triangle QST and RTS are congruent.
By CPCT,
Since the sum of two consecutive angles of a parallelogram is 180 degree.
Therefore the measure of angle S and angle T are 90 degree.
Since the measure of all interior angles of QRST are 90 degree, therefore the parallelogram QRST is a rectangle.
F = 1.8C + 32
С = -10 ⇒ F = 1.8·(-10) + 32 = 14 ⇒ (-10, 14)
С = 0 ⇒ F = 1.8·(0) + 32 = 32 ⇒ (0, 32)
С = 10 ⇒ F = 1.8·(10) + 32 = 50 ⇒ (10, 50)
С = 20 ⇒ F = 1.8·(20) + 32 = 68 ⇒ (20, 68)
С = -10 ⇒ F = 1.8·(-10) + 32 = 14 ⇒ (-10, 14)
С = 0 ⇒ F = 1.8·(0) + 32 = 32 ⇒ (0, 32)
С = 10 ⇒ F = 1.8·(10) + 32 = 50 ⇒ (10, 50)
С = 20 ⇒ F = 1.8·(20) + 32 = 68 ⇒ (20, 68)
