A body is forced by F 1 = 6N to produce an acceleration of a 1 = 3m / S 2. How much force should be applied to make a 2 = 9m / S 2 acceleration

A body is forced by F 1 = 6N to produce an acceleration of a 1 = 3m / S 2. How much force should be applied to make a 2 = 9m / S 2 acceleration

F1=ma1
m=F1/a1
F2=ma2=a2F1/a1=9x6/3=18N

If the function f (x) = x2-bx + C satisfies f (1 + x) = f (1-x) and f (0) = 3, then the relationship between F (BX) and f (Cx) is ()
A. F (BX) ≤ f (Cx) B. f (BX) ≥ f (Cx) C. f (BX) > F (Cx) d

∵ f (1 + x) = f (1-x), the axis of symmetry of ∵ f (x) image is a straight line x = 1, thus B = 2. F (0) = 3, ∵ C = 3. ∵ f (x) decreases on (- ∞, 1) and increases on (1, + ∞). If x ≥ 0, then 3x ≥ 2x ≥ 1, ∵ f (3x) ≥ f (2x). If x < 0, then 3x < 2x < 1, ∵ f (3x) > F (2x). ∵ f (3x) ≥ f (2x)

Y = x ^ / 1 + x ^, find the value of F (m) + F (1 / M), f (1) + F2 + F3 + F4 + F1 / 2 + F1 / 3 + F1 / 4

f(m)+f(1/m)=m^2/(1+m^2)+(1/m^2)/(1+(1/m)^2)=m^2/(1+m^2)+1/(1+m^2)=1
f(1)+f2+f3+f4+f1/2+f1/3+f1/4=f1+f1+f2+f1/2+f3+f1/3+f4+f1/4=1+1+1+1=4

The three forces f1f2f3 acting on a point in the plane are in equilibrium, | F1 | = 1n, | F2 | = (√ 6 + √ 2) / 2n, and the angle between F1 and F2 is 45 ° to find the size of F3
The formula that uses Pythagorean theorem needs detailed process
This is a mathematical problem, about the vector, the answer is the root 3 + 1, I just want how to use Pythagorean to calculate the answer

Since the three forces are balanced, it must satisfy that the vector graph of the three forces can form a triangle. As a triangle, there are two side lengths and included angles. The first choice is the cosine theorem. I don't know if you have learned it or not? This problem can't use the Pythagorean theorem, because I don't know if it is a right triangle ∵ (√ 6 + √ 2) / 2 > 1 ∵