Answer:
|F net| = 20.22 N
θ ≈ 19.8°
Step-by-step explanation:
F net = 15N i + 8cos(60°)N i + 8sin(60°)N j
= 15N i + 8×½N i + 8×√3/2N j
= 15N i + 4N i + 4√3N j
= 19N i + 4√3N j
|F net| = √(19²+(4√3)²) = √(361+48) = √409 ≈ 20.22N
tan(θ) = 4√3 ÷ 19 ≈ 0.36 → θ ≈ arctan(0.36) = 19.8°
Suppose
At the same time, we can write
Note that (just reverse the sum). Let's pair the first terms of and , and the second, and the third, and so on:
Now, each grouped term in the sum on the right side adds to 151. There are 52 grouped terms on that same side (because there are 50 numbers in the range of integers 51-100, plus 50 and 101), which menas
But , as we pointed out, so
Answer:
4'9.87
Step-by-step explanation:
Answer:
150 to 117 is 25%
340 to 255 is 25%
450 to 351 is 22%
Step-by-step explanation: