The coefficient is 9 because it is not talking about the variable or power.
5) The relation between intensity and current appears linear for intensity of 300 or more (current = intensity/10). For intensity of 150, current is less than that linear relation would predict. This seems to support the notion that current will go to zero for zero intensity. Current might even be negative for zero intensity since the line through the points (300, 30) and (150, 10) will have a negative intercept (-10) when current is zero.
Usually, we expect no output from a power-translating device when there is no input, so we expect current = 0 when intensity = 0.
6) We have no reason to believe the linear relation will not continue to hold for values of intensity near those already shown. We expect the current to be 100 for in intensity of 1000.
8) Apparently, times were only measured for 1, 3, 6, 8, and 12 laps. The author of the graph did not want to extrapolate beyond the data collected--a reasonable choice.
The answer is C. X^2-6x+9
The perimeter of ΔXYZ is 126 units.
Solution:
Given ΔPQR
ΔXYZ.
In ΔPQR,
PQ = 5, QR = 10, PR = 6
In ΔXYZ, XY = 30
Perimeter of ΔPQR = PQ + QR + PR
= 5 + 10 + 6
= 21
Perimeter of ΔPQR = 21
To find the perimeter of ΔZYZ:
If two triangles are similar then the ratio of the perimeters of two similar triangles is same as the ratio of their corresponding sides.


Do cross multiplication, we get
⇒ 5 × Perimeter of ΔXYZ = 30 × 21
⇒ 5 × Perimeter of ΔXYZ = 630
Divide by 5 on both sides of the equation.
⇒ Perimeter of ΔXYZ = 126
Hence the perimeter of ΔXYZ is 126 units.
Answer:
Volume of cone = 200.96 square feet
Step-by-step explanation:
We are given the following information in the question:
The tree is cone shaped and has a radius of 4 feet.
r = 4 feet
The height of the tree is three times its radius.

Volume of tree =

Putting all the values, we get,

Radius of cone whose height is is equal to three times its radius is given as:
![V = \displaystyle\frac{1}{3}\pi r^2 h\\\\V = \displaystyle\frac{1}{3}\pi r^2 (3r)\\\\3V = 3\pi r^3\\\\ r = \sqrt[3]{\frac{3V}{3\pi}}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cdisplaystyle%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2%20h%5C%5C%5C%5CV%20%3D%20%20%5Cdisplaystyle%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2%20%283r%29%5C%5C%5C%5C3V%20%3D%203%5Cpi%20r%5E3%5C%5C%5C%5C%20r%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B3%5Cpi%7D%7D)