Can u post a clearer pic? i can help u if u do :)
Answer:
Part a) 
Part b) 
Step-by-step explanation:
Part a)
Let
x -----> the number of weeks
y ----> the total amount Chem has deposited in a saving account
we know that
The equation of a line in slope intercept form is
where
m is the rate or slope of the linear equation
b is the y-intercept of the linear equation or initial value (value of y when the value of x is equal to zero)
In this problem we have that
The rate or slope is equal to
-----> amount deposited by Chem each week
The y-intercept or initial value is
----> amount deposited originally in the saving account
substitute the values
Part b) How much has Chem deposited 30 weeks after his initial deposit?
For x=30 weeks
substitute in the equation
<span>6554
Assuming that the growth the function will be of the form
f(n) = P*R^n
where
P = Initial population
R = Growth rate per period
n = number of periods
In this problem, the only period or interval I see are days. So we'll use n to represent the number of days since the start. The initial population is 1000, and the ratio would be 1600/1000 = 1.6. So the formula becomes
f(n) = 1000*1.6^n
So let's evaluate that function using n=4.
f(n) = 1000*1.6^n
f(4) = 1000*1.6^4
f(4) = 1000*6.5536
f(4) = 6553.6
Since we can't have a fractional mosquito, round to the nearest integer, giving 6554.</span>
Answer:
Below in bold.
Step-by-step explanation:
a) the curve moves up 5 units, so
the vertex is (2, -3 + 5)
= (2, 2).
b) - f(x) is the image of f(x) in the x-axis so
The vertex is at (2, 3).
