Answer:
![y = \frac{1}{6} x](https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B1%7D%7B6%7D%20x)
Step-by-step explanation:
The equation of a line can be written in the form of y=mx+c, where m is the gradient and c is the y-intercept.
![\boxed{gradient = \frac{y1 - y2}{x1 - x2} }](https://tex.z-dn.net/?f=%5Cboxed%7Bgradient%20%3D%20%20%5Cfrac%7By1%20-%20y2%7D%7Bx1%20-%20x2%7D%20%7D)
Using the above formula,
![m = \frac{1 - ( - 1)}{6 - ( - 6)} \\ m = \frac{1 + 1}{6 + 6} \\ m = \frac{2}{12} \\ m = \frac{1}{6}](https://tex.z-dn.net/?f=m%20%3D%20%20%5Cfrac%7B1%20-%20%28%20-%201%29%7D%7B6%20-%20%28%20-%206%29%7D%20%20%5C%5C%20m%20%3D%20%20%5Cfrac%7B1%20%2B%201%7D%7B6%20%2B%206%7D%20%20%5C%5C%20m%20%3D%20%20%5Cfrac%7B2%7D%7B12%7D%20%20%5C%5C%20m%20%3D%20%20%5Cfrac%7B1%7D%7B6%7D%20)
Substitute the value of m into the equation:
![y = \frac{1}{6} x + c](https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B1%7D%7B6%7D%20x%20%2B%20c)
To find the value of c, substitute a pair of coordinates.
When x=6, y=1,
![1 = \frac{1}{6} (6) + c \\ 1 = 1 + c \\ c = 1 - 1 \\ c = 0](https://tex.z-dn.net/?f=1%20%3D%20%20%5Cfrac%7B1%7D%7B6%7D%20%286%29%20%2B%20c%20%5C%5C%201%20%3D%201%20%2B%20c%20%5C%5C%20c%20%3D%201%20-%201%20%5C%5C%20c%20%3D%200)
Thus, the equation of the line is y=⅙x.
Answer:
in scientific notation....14×10^12miles
Answer:
i'd say no. the first hour he earned $20 but hour 2 is only a $10 increase...and so on...
Answer:
58%
Step-by-step explanation:
The first thing that you have to do if figure out how many sandwiches are left we do this by the total - bad sandwiches = good sandwiches
12-5=7
There are 7 good sandwiches left. To figure out the percentage you do
=
now we fill in the parts the part is 7, the total is 12 and the % is x
=
now we cross multiply, 7 x 100 and 12 x X
700=12x now we divide both sides by 12 so the x is alone
x = 58.33 now we round down because it ends in.3
x=58
Answer:
After 80 years 432 pounds of radioactive element will be remaining.
Step-by-step explanation:
A function
models the amount of a radioactive material.
Here x = number of years taken for decay.
If
pound and
pound, then we have to calculate the time for decay.
![A_{t}=A_{0}e^{-0.0077x}](https://tex.z-dn.net/?f=A_%7Bt%7D%3DA_%7B0%7De%5E%7B-0.0077x%7D)
![e^{-0.0077x}=\frac{432}{800}](https://tex.z-dn.net/?f=e%5E%7B-0.0077x%7D%3D%5Cfrac%7B432%7D%7B800%7D)
Take log on both the sides of the equation
![ln(e^{-0.0077x})=ln(\frac{432}{800} )](https://tex.z-dn.net/?f=ln%28e%5E%7B-0.0077x%7D%29%3Dln%28%5Cfrac%7B432%7D%7B800%7D%20%29)
-0.0077x = ln(0.54)
-0.0077x = -0.616186
x = ![\frac{0.616186}{0.0077}](https://tex.z-dn.net/?f=%5Cfrac%7B0.616186%7D%7B0.0077%7D)
x = 80.02 years
Therefore, After 80 years 432 pounds of radioactive element will be remaining.