So u will do 88+84=172+88=260+91=351 and that is all four tests added up now u have to divide the answer by 4 because it was four tests averages
351/4=87.75 wich is aproxamately 88
so with all four of them added wich is 351 u will add the answer choices so 351+96=447/5=89.4 wich is aproxamately 89
351+97=448/5=89.6 wich is aproxamately 90
351+98=449/5=89.8 wich is aproxamately 90
351+99=450/5=90
so i would say that 99 points is the one she needs to score because it isnt asking for aproxamately
so it is 99 points
Answer:
Question 11.1
At 7 : 00 pm in Sydney, it will be 10 : 00 pm in Berlin.
Question 11.2
Place | Time
Sydney | 5 : 00 pm
Berlin | 8 : 00 pm
It would be a good time for Mark and Hans to cha,t when it's 5 : 00 pm in Sydney and 8 : 00 pm in Berlin.
hope that helps ...
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
Answer:
Step-by-step explanation:
-4
Answer:
Step-by-step explanation:
pull like terms from the problem to re-arrange it into a product
