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Minchanka [31]
2 years ago
15

Find the exact value of the trigonometric function given that sinU=-7/25 and cosV=-4/5 (Both U and V are in Quadrant III)

Mathematics
1 answer:
Contact [7]2 years ago
3 0

cosU:-

\\ \rm\longmapsto \sqrt{1-sin^2U}=\sqrt{1-49/625}=24/25

As U lies in Q3

  • cosU=-24/25

sinV

\\ \rm\longmapsto \sqrt{1-cos^2V}=\sqrt{1-16/25}=3/4

As V lies in Q3

  • sinV=-3/5

So

  • sin(V-U)=sinVcosU-cosVsinU=(-3/5)(-24/25)-(-4/5)(-7/25)=72/125-28/125=72-28/125=<u>4</u><u>4</u><u>/</u><u>1</u><u>2</u><u>5</u>
  • cos(U-V)=cosUcosV+sinUsinV=(-24/25)(-4/5)+(-7/25)(-3/5)=96/125+21/125=117/125
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Drew jumped farther than all 4 students above but jumped shorter than 7 feet , 7 inches. How far could drew have jumped? Write s
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<h3>The distance covered by Drew in long jump is either 7 ft 5 inches OR 7 ft  6 inches.</h3>

Step-by-step explanation:

Here, the needed table is attached for the reference.

Now, as we can see from the table:

The distance covered by  Cindy  = 2 yards, 1 foot 3 inches

The distance covered by  Tyrette   =  7 feet, 2 inches

The distance covered by Nina   =   2 yards, 1 foot, 1 inch

The distance covered by  Monique   = 7 feet, 4 inches

As we know : 1 yard  =  3 feet

So, the distance covered  by :

Cindy  =  2 yards, 1 foot 3 inches = 3 ft x (2)  + 1 ft +  3 in  = 7 ft 3 in

Tyrette   =  7 feet, 2 inches

Nina   =   2 yards, 1 foot, 1 inch  = = 3 ft x (2)  + 1 ft +  1 in  = 7 ft 1 in

Monique   = 7 feet, 4 inches

So by comparing all distances, we can see that:

The maximum distances jumped by all four  is 7 ft 4 inches.

The distance covered by Drew is less than 7 ft 7 in.

So, he must have jumped 7 ft 5 inches OR 7 ft  6 inches.

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Answer:

The equation simplified equals to 7

Step-by-step explanation:

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