Answer:
Degree 4
Step-by-step explanation:
Expression:
3x^3 + 3x^2y^2 - y^2
- 3x^3 - degree 3
- 3x^2y^2 - degree 2+2 = 4
- y^2 - degree 2
So the expression has degree 4 as per the highest degree of the terms
9514 1404 393
Answer:
$9072
Step-by-step explanation:
Put the given numbers in the given formula and do the arithmetic.
A = MN . . . . . . M = payment = $189. N = number of payments = 48.
A = $189 × 48 = $9072
The total amount paid is $9072.
Answer:
n = 0
Step-by-step explanation:
Multiply parenthesis by 3 and get:
1.) 18 + 9n = 5n + 18 + 3n
Then cancel the equal terms:
9n = 5n + 3n
Collect like terms:
9n = 8n
Move variable to the left:
9n - 8n = 0
n + 0
Plugin the value of 0 into x:
3 (6 + 3 x 0) = 18
5 x 0 + 3 x 0 + 18 = 18
*If the plugged in value have the same result than the answer is correct because 18 + 18.
Answer:
2:25pm
Step-by-step explanation:
The time given is 5pm to catch the train back to school;
Amount of time for the tour = 1hr 45min
= 105min
Amount of time at gift shop = 30min
Amount of time to save = 20min
Total time = 105min + 30min + 20min = 155min
So;
Total time = 2hr 35min
So;
5pm - 2hr 35min = 2:25pm
Answer and Explanation:
A function is said to be increasing, if the derivative of function is f’(x) > 0 on each point. A function is said to be decreasing if f”(x) < 0.
Let y = v (z) be differentiable on the interval (a, b). If two points z1 and z2 belongs to the interval (a, b) such that z1 < z2, then v (z1) ≤ v (z2), the function is increasing in this interval.
Similarly, the function y = v(z) is said to be decreasing, when it is differentiable on the interval (a , b).
Two points z1 and z2 Є (a, b) such that z1 > z2, then v (z1) ≥ v(z2). The function is decreasing on this interval.
The function y = v (z)
The derivative of function Y’ = v’(z) is positive, then the function is increasing.
The function y = v (z)
The derivative of function y’ is negative, then the function is decreasing.
