T = k/R where k is the constant of variation, T=time and R=speed.
Plug in the given values:-
12 = k/60
k = 12*60 = 720
So the equation of variation is T = 720/R
At 85 mph we have
T = 720 / 85 = 8.47 hours (answer)
Answer:
The unit rate is 0.11 $/ounce
Step-by-step explanation:
we know that
To find the unit rates divide the total cost by the total volume
so
(6/54)=0.11 $/ounce
Answer:
2.1
Step-by-step explanation:
Answer: 200 square units
Step-by-step explanation:
Given
Circumference of the circle is ![20\pi](https://tex.z-dn.net/?f=20%5Cpi)
Suppose r is the radius of the circle
The biggest area of a quadrilateral that can fit in a circle is of square.
Deduce the radius of the circle
![2\pi r=20\pi\\r=10\ \text{units}](https://tex.z-dn.net/?f=2%5Cpi%20r%3D20%5Cpi%5C%5Cr%3D10%5C%20%5Ctext%7Bunits%7D)
Suppose the side of the square is a
from the figure, we can write
![\Rightarrow a^2+a^2=(2r)^2\\\Rightarrow 2a^2=4r^2\\\Rightarrow a=\sqrt{2}r\\\Rightarrow a=10\sqrt{2}\ \text{units}](https://tex.z-dn.net/?f=%5CRightarrow%20a%5E2%2Ba%5E2%3D%282r%29%5E2%5C%5C%5CRightarrow%202a%5E2%3D4r%5E2%5C%5C%5CRightarrow%20a%3D%5Csqrt%7B2%7Dr%5C%5C%5CRightarrow%20a%3D10%5Csqrt%7B2%7D%5C%20%5Ctext%7Bunits%7D)
Area of the quadrilateral is
![\Rightarrow A=(10\sqrt{2})^2\\\Rightarrow A=100\times 2\\\Rightarrow A=200\ \text{square units}](https://tex.z-dn.net/?f=%5CRightarrow%20A%3D%2810%5Csqrt%7B2%7D%29%5E2%5C%5C%5CRightarrow%20A%3D100%5Ctimes%202%5C%5C%5CRightarrow%20A%3D200%5C%20%5Ctext%7Bsquare%20units%7D)