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Arada [10]
1 year ago
7

Can you please help me with this i don’t understand

Mathematics
1 answer:
Alinara [238K]1 year ago
6 0

Answer:

  (5, 2)

Step-by-step explanation:

<h3>understanding the question</h3>

The question is asking you to express the temperature at 5 pm as a pair of graph coordinates.

Doing this requires that you ...

  • read from the graph the temperature at 2 pm
  • read from the graph the temperature at 8 pm
  • figure the temperature halfway between those values
  • express the 5 pm temperature as an ordered pair (hours, °C)

__

<h3>reading the graph</h3>

The description of the graph tells you the horizontal scale is hours after noon, so the temperature at 2 pm will be a point on the graph with an x-coordinate of 2. That point is (2, 7), indicating the temperature at 2 pm is 7 °C.

Likewise, the temperature at 8 pm is shown as the point on the graph with x-coordinate 8. That point is (8, -3), indicating a temperature of -3°C at 8 pm.

<h3>halfway temperature</h3>

The temperature halfway between those values (7 °C and -3 °C) can be found several ways. Perhaps the easiest is to average the values:

  (7 +(-3))/2 = 4/2 = 2

The temperature at 5 pm is 2 °C.

<h3>coordinates</h3>

Coordinates on the graph are ordered pairs of the form (x, y) = (hours, °C). The time 5 pm is 5 hours after noon, so the x-coordinate is 5. The temperature at 5 pm is halfway between the temperatures at 2 and 8 pm, which we have computed to be 2 °C, so the y-coordinate is 2.

The coordinates that represent the temperature at 5 pm are ...

  (5, 2)

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Answer:

\theta_{CAB}=128.316

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we can use the cosine law to find the angle:

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the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}

\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}

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Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}

\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)

\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)

\theta_{ABC}=\arcsin{0.4359}\right)

\theta_{ABC}=25.842

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

\theta_{BCA}=25.842

this can also be checked using the fact the sum of all angles inside a triangle is 180

\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180

25.842+128.316+25.842

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