Answer:
for r= 70 cm
the rate is 0.008cm/s
for r=95cm
the rate= 0.0044cm/s
Step-by-step explanation:
kindly find attached a rough sketch of the circle and a detailed solution of the problem.
Given data
rate dv/dt=500cm^3
r=70cm
r=95cm
Answer:
y + 1 = -14(x - 2)^2y = -14(x - 2)^2 - 1
Step-by-step explanation:
You do 65+90=155
then 180-155=25 to get the angle next to w
w=155
Answer:
As per the statement:
The path that the object takes as it falls to the ground can be modeled by:
h =-16t^2 + 80t + 300
where
h is the height of the objects and
t is the time (in seconds)
At t = 0 , h = 300 ft
When the objects hit the ground, h = 0
then;
-16t^2+80t+300=0
For a quadratic equation: ax^2+bx+c=0 ......[1]
the solution for the equation is given by:
On comparing the given equation with [1] we have;
a = -16 ,b = 80 and c = 300
then;
Simplify:
= -2.5 sec and = 7.5 sec
Time can't be in negative;
therefore, the time it took the object to hit the ground is 7.5 sec
Answer:
Step-by-step explanation:
I attached an image to aid the understanding of the question.
Looking at the image, we see that the 8 shaded parts are congruent, as affirmed in the question as well. And we are told that T = 3, this implies that the area of the square with T as it's side is 9ft². Since all the 8 squares are congruenrt, it means each square has its area to be 9. Therefore, the total area of the 8 shaded squares will be:
It remains the area of the shaded square with side S.
From the question, we have the following ratio:
I did not add ± because length is always positive, so the case of negative is eliminated.
Now the areas of S is
Therefore, the total area of the shaded squares is