Lna^b = blna
let y = lne^e
so y = lne^e = e lne
lne = 1
y = e which is a constant (Euler's number) ~= 2.718
For Euler's number estimating methods request in a comment.
Answer:
_______________________________________________________
The rewritten equation is:
" - 3x² + 30 + 225 = 0 " ;
which is ultimately simplified to:
" x² − 10 − 75 = 0 " .
___________________________________________________
The number of days is: "15 days" .
___________________________________________________<span>
Explanation:
______________________________________________
Given:
h(x) = - 3x</span>² <span>+ 30 + 225 ;
Rewrite as: " </span>- 3x² + 30 + 225 = 0 " ;
__________________________________________________________
Multiply EACH SIDE of the equation by "-1" ; to get rid of the "-3" ;
-1 * {- 3x² + 30 + 225 = 0} ;
to get: " 3x² − 30 <span>− 225 = 0" ;
Divide EACH SIDE of the equation by "3" ; to simplify; since:
"3", "30", and "225" are divisible by "3" ;
{</span>3x² − 30 − 225} / 3 = 0 / 3 ;
to get:
______________________________________
" x² − 10 − 75 = 0 " ;
______________________________________
Let us factor:
-15 + 5 = -10 ;
-15 * 5 = -75 ;
_______________________________________
Rewrite as:
(x − 15) (x + 5) = 0 ;
x = 15; and x = -5 ;
_______________________________________
The number of days cannot be "negative value"; so we stick with:
"x = 15 days"
___________________________________________________
Answer: The rewritten equation is:
" - 3x² + 30 + 225 = 0 " ;
which is ultimately simplified to:
" x² − 10 − 75 = 0 " .
___________________________________________________
The number of days is: "15 days" .
___________________________________________________
Answer:
not possible to answer on text, very sorry we cant help
Answer:
25-50
Step-by-step explanation:
Since the graph shows us only values of x that are between 25 and 50, this is the correct answer.
We can interpolate with known data. However, once the interpolation occurs, and we approximate the data for a line of the form <em>y = mx + b</em> (m <0 , linear negative association), we could find the other values of y for a wider range of x values.