At the beginning of year 1, Bode invests $250 at an annual simple interest rate of 3%. He makes no deposits to or withdrawals fr
1 answer:
Answer:
D.A(n) = 250 + (n – 1)(0.03 • 250); $347.50
Step-by-step explanation:
we know that
The simple interest formula is equal to
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
in this problem we have
substitute in the formula above
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Answer:
An equation in the slope-intercept form is:
y = a*x + b
Where a is the slope, and b is the y-intercept.
a)
Here we have a slope of 6 and a y-intercept of -3
Then the equation is:
y = 6*x - 3
Now we want to graph this.
To graph it, we first need to find two points (x, y) that belong to this equation, then we can graph the points, and connect them with a line.
To find the points, we evaluate in two different values of x.
x = 0
y = 6*0 - 3 = -3
Then we have the point (0, -3)
x = 1
y = 6*1 - 3 = 3
Then we have the point (1, 3)
The graph of this line can be seen in the image below (the red one)
b) Similar to before, here the slope is -3/5, then the equation is something like:
y = (-3/5)*x + b
Now we also know that the line passes through the point (-10, 8)
This means that when x = -10, we must have y = 8
Replacing these two in the equation we get:
8 = (-3/5)*-10 + b
8 = 6 + b
8 - 6 = 2 = b
Then this equation is:
y = (-3/5)*X + 2
The graph can be found in the same way as before, the graph of this function can also be seen in the image below (the green one)
