Let's find the rates at which Melinda and Marcus are saving money and compare them.
Based on the table, we can see that Melinda originally had $75. Then she got $135 in 5 weeks. So the rate of saving money is (135–75)/5 = $12 per week.
This rate is unchanged for the next weeks. As we can see, she got $195 in the next 5 weeks. So she saved $60 more those 5 weeks, or the rate is $60/5 = $12 per week again.
So Melinda saved $12 per week.
As for Marcus, the equation tells us that the rate of saving money is $14 which is the coefficient in front of x.
Hence, t<span>he rate at which Melinda is adding to her savings each week is 2$
less than the rate at which Marcus is adding to his savings each week.</span>
Answer: b
Step-by-step explanation:
Answer:

Step-by-step explanation:
Consider a sketch of the problem as shown in the picture, where:
- Blue line is given by y = 4x + 1.
- Point B is the center of the circle.
- Point A is (-3, 0).
Since the center of the circle lies on the line y = 4x +1 and is tangent to the x-axis at point A, then its radius BA is perpendicular to the x-axis. To find the coordinates of point B, we must replace x = -3 into the blue line equation: y = 4x(-3) + 1 = -11.
So, we know that the center of the circle is at B=(-3, -11). And furthermore, the radius BA is of length r=11.
Since the <em>general equation of the circle</em> of radius lenght r centered at (h, k) is given by

then with h = -3, k = -11 and r= 11, the equation of our circle is

Answer:
6 + (d * 2)
Step-by-step explanation:
You are adding 6 to the equation after the amount d times 2. In that case, a parentheses is required to round up d * 2.
Answer:
Yes, this graph is absolutely right for the function y =
.
Step-by-step explanation:
firstly we will calculate amplitude, period, phase shift, and vertical shift.
Use the standard form
to calculate the values of amplitude, period, phase shift, vertical shift.
here a = 2
b = 2
c = 0
d = -3
Amplitude = |a| = 2
period =
=
= 
phase shift =
= 0
and vertical shift = d = -3
so, by using this information, if we plot the graph then the graph obtained will be same as the given graph.