Answer:
see explanation
Step-by-step explanation:
The translation represented by ![\left[\begin{array}{ccc}1\\4\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C4%5C%5C%5Cend%7Barray%7D%5Cright%5D)
interprets as a shift of 1 unit to the right ( add 1 to x- coordinate ) and a
shift of 4 units down ( subtract 4 from the y- coordinate ), then
(1, 4 ) → (1 + 1, 4 - 4 ) → (2, 0 )
(4, 4 ) → (4 + 1, 4 - 4 ) → (5, 0 )
(6, 2 ) → (6 + 1, 2 - 4 ) → (7, - 2 )
(1, 2 ) → (1 + 1, 2 - 4 ) → (2, - 2 )
Answer:
4
Step-by-step explanation:
Make the 10% a decimal
.10
Then multiply it by 40
.10*40
You get 4
Answer:
.22 - .06= .16
.16y should be your equation
Answer:
D 3.5
Step-by-step explanation:
Replace values in the equation with values given. Calculate.
Answer:
So it will take 15 year from 1998 to reach population 229 thousands.
i.e. In 2013 population will be 229 thousand.
Step-by-step explanation:
Given:
Formula model population for city A=146e0.03t
To Find:
Years required to reach population 229 thousands after 1998
Solution:
Using the model ,
A=146e0.03t
here A=229
So equation becomes,
229=146e0.03t
(229/146)=e0.03t
Using log function with base 10 on both sides we get ,
Log(229/146)=log(e^0.03t)
log(1.56489)=0.03tlog(e) .................(using log property log(a^n)=nloga)
0.19548=0.03*t*0.43429
0.19548/0.013028=t
t=(0.19548/0.013028)
t=15.01
t=15 years
So it will take 15 year from 1998 to reach population 229 thousands.