Y=-2 is a horizontal line, each point which lies on this line has the y-coordinate -2.
Only in option C both points have the y-coordinate equal to -2.
The answer is C.
Answer:
1. C. cylindrical coordinates
2 A. spherical coordinates
3. A. spherical coordinates
4. D. Cartesian coordinates
5 B. polar coordinates
Step-by-step explanation:
USE THE BOUNDARY INTERVALS TO IDENTIFY
1. ∭E dV where E is:
x^2 + y^2 + z^2<= 4, x>= 0, y>= 0, z>= 0 -- This is A CYLINDRICAL COORDINATES SINCE x>= 0, y>= 0, z>= 0
2. ∭E z^2 dV where E is:
-2 <= z <= 2,1 <= x^ 2 + y^2 <= 2 This is A SPHERICAL COORDINATES
3. ∭E z dV where E is:
1 <= x <= 2, 3<= y <= 4,5 <= z <= 6 -- This is A SPHERICAL COORDINATES
4. ∫10∫y^20 1/x dx ---- This is A CARTESIAN COORDINATES
5. ∬D 1/x^2 + y^2 dA where D is: x^2 + y^2 <=4 This is A POLAR COORDINATES
Answer:
2,033-993.50= 1039.50/74.25= 14
Step-by-step explanation:
not sure the exact equation but p would be 14 and you'd do something similar to that shown above
A is correct because 445.50 Divided By 150 is Equal to $2.97.
Answer:
Part a) 
Part b) 
Step-by-step explanation:
Part a) Write an equation for T (d)
Let
d ----> the number of days
T ---> the time in minutes of the treadmill
we know that
The linear equation in slope intercept form is equal to
where
m is the slope or unit rate
b is the y-intercept or initial value
In this problem we have
The slope or unit rate is
The y-intercept or initial value is
substitute
Part b) Find T (6), the minutes he will spend on the treadmill on day 6
For d=6
substitute in the equation and solve for T
