X=14
How to:
Square both sides then solve the equation
Answer:
for example, if f(x)= x^2 + 3
and they gave write f(5)= 5^2 +3
=25+3=28
5)
a. The equation that describes the forces which act in the x-direction:
<span> Fx = 200 * cos 30 </span>
<span>
b. The equation which describes the forces which act in the y-direction: </span>
<span> Fy = 200 * sin 30 </span>
<span>c. The x and y components of the force of tension: </span>
<span> Tx = Fx = 200 * cos 30 </span>
<span> Ty = Fy = 200 * sin 30 </span>
d.<span>Since desk does not budge, </span><span>frictional force = Fx
= 200 * cos 30 </span>
<span> Normal force </span><span>= 50 * g - Fy
= 50 g - 200 * sin 30
</span>____________________________________________________________
6)<span> Let F_net = 0</span>
a. The equation that describes the forces which act in the x-direction:
(200N)cos(30) - F_s = 0
b. The equation that describes the forces which act in the y-direction:
F_N - (200N)sin(30) - mg = 0
c. The values of friction and normal forces will be:
Friction force= (200N)cos(30),
The Normal force is not 490N in either case...
Case 1 (pulling up)
F_N = mg - (200N)sin(30) = 50g - 100N = 390N
Case 2 (pushing down)
F_N = mg + (200N)sin(30) = 50g + 100N = 590N
binomial(16 + 7, 16) 2^(-(16 + 7)) = ((16 + 7)!)/(16! 7! 2^(16 + 7)) = 245157/8388608 ≈ 0.02922 ≈ 1/34.22
(assuming children are independent and male and female are equally likely)
| probability
less than 16 boys | 0.9534
16 or less boys | 0.9827
more than 16 boys | 0.01734
16 or more boys | 0.04657
fraction of boys | 16/(16 + 7) ≈ 0.695652
fraction of girls | 7/(16 + 7) ≈ 0.304348
expected value | 11.5
standard deviation | 2.398
variance | 5.75
11.5
The question is asking which formula will give the results
1, 9, 36, 100, 225, ...
for k values of
1, 2, 3, 4, 5, ...
You can try them out to see.
A) for k=2, gives 2*3/2 = 3 . . . not 9
B) for k=1, gives 2^3/3 = 8/3 . . . not 1
C) for k=2, gives (2^2*3^2)/4 = 9 . . . . looks promising
D) for k=1, gives 1*2^3/5 = 8/5 . . . not 1
Selection C is the only viable choice. (And the correct one.)
