Answer:
C
Step-by-step explanation:
(3.6xy^5)(2.5x^2y^-2)
9x^3y^3
Answer:
option 4 ⇒ 236 lb.
Step-by-step explanation:
Best explanation of the question is as shown in the attached figure.
we will use the parallelogram method to calculate resultant force.
to get the length of the resultant force ⇒ use the cosines law
The cosine law is a² = b² + c² - 2 * b * c * cos (∠A)
Applying at the question where b = F₁ , c = F₂ and ∠A = ∠x
Given that F₁ = 100 pounds , F₂ = 150 pounds and ∠x = 180° - 40° = 140°
∴ (Resultant force)² = 100² + 150² - 2 * 100 * 150 * cos (∠140) = 55481
∴ Resultant force = √55481 = 235.54 ≅ 236 pounds
The answer is option 4 ⇒ 236 lb.
Answer:
I can't help.
Step-by-step explanation:
There is no question, so therefore, I can not help.
With the curve

parameterized by

with

, and given the vector field

the work done by

on a particle moving on along

is given by the line integral

where

The integral is then


Answer:
Opps i am really sorry i donot know it