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noname [10]
3 years ago
14

The table shows the temperatures in degrees Celsius, y, inside a shed at x hours past noon. The maximum and minimum temperatures

are shown.
Which equation models this data?

Mathematics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

B

y = 3 cos (pi/12x - pi/4) + 17

Step-by-step explanation:

First find the average temperature which is 20 + 14/2 = 17

Find the amplitude

20 - 17 = 3

This means that that at noon, the shift is 3hrs past

the best equation that represents this scenario is y = 3 cos (\frac{\pi}{12}x - \frac{\pi}{4}) + 17

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I would factor of -1

Step-by-step explanation:

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3 years ago
Is the product of fifteen negative integers positive or negative?
enot [183]

Answer:

the sign of the product of 15 negative numbers will be negative

Step-by-step explanation:

Heya mate!!

The product of odd numbers is odd and the product of even numbers is even.

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3 years ago
SOLVED
Nastasia [14]
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
4 0
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