I'm assuming that the diagonals intersect at E. Since diagonals of a parallelogram bisect each other, BE=ED.
- 7x-2=x²-10
- x²-7x-8=0
- (x-8)(x+1)=0
- x = -1, 8
As distance must be positive, we reject the negative case, so x=8.
Thus, BE=ED=54.
Consider, in ΔRPQ,
RP = R (Radius of larger circle)
PQ = r (radius of smaller circle)
We have to find, RQ, by Pythagoras theorem,
RP² = PQ²+RQ²
R² = r²+RQ²
RQ² = R²-r²
RQ = √(R²-r²
Now, as RQ & QS both are tangents of the smaller circle, their lengths must be equal. so, RS = 2 × RQ
RS = 2√(R²-r²)
Answer:
A
Step-by-step explanation:
This is exponential decay; the height of the ball is decreasing exponentially with each successive drop. It's not going down at a steady rate. If it was, this would be linear. But gravity doesn't work on things that way. If the ball was thrown up into the air, it would be parabolic; if the ball is dropped, the bounces are exponentially dropping in height. The form of this equation is
, or in our case:
, where
a is the initial height of the ball and
b is the decimal amount the bounce decreases each time. For us:
a = 1.5 and
b = .74
Filling in,

If ww want the height of the 6th bounce, n = 6. Filling that into the equation we already wrote for our model:
which of course simplifies to
which simplifies to

So the height of the ball is that product.
A(6) = .33 cm
A is your answer
Answer:
C. None of the above
Step-by-step explanation: