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LiRa [457]
3 years ago
8

Cos^2 2A+4sin^2A×cos^2 A=1

Mathematics
1 answer:
Nonamiya [84]3 years ago
3 0
Using Sin^2 and Cos^2 identities,
= Cos^2(2A) + 4[(1/2)(1-Cos(2A)][(1/2)(1+Cos(2A)] = 1
= Cos^2(2A) + 1 -Cos^2(2A) = 1
   0 = 0
There is nothing to solve for as it is an identity of sorts.

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Point (2, 4) is reflected across the x-axis. What are <br> the coordinates of its image?
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Answer:

The coordinates of its image are (2, -4)

Step-by-step explanation:

<em>Let us revise the rules of reflection across the axes</em>

  • If the point (x, y) reflected across the x-axis, then its image is (x, -y), the rule of reflection is rx-axis (x, y) → (x, -y)
  • If the point (x, y) reflected across the y-axis, then its image is (-x, y), the rule of reflection is ry-axis (x, y) → (-x, y)

∵ The point (2, 4) is reflected across the x-axis

→ By using the first rule above rx-axis (x, y) → (x, -y)

∴ Change the sign of its y-coordinate

∴ Its image is (2, -4)

∴ The coordinates of its image are (2, -4)

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3 years ago
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Answer:

200X = Y

Step-by-step explanation:

Since the insect in question can flap its wings 200 times per second, the equation that represents how many times it flaps its wings according to the time of flight is the following:

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Thus, 200 is the number of times per second it flaps its wings, while X is the number of seconds in question. In turn, Y is the number of times it flaps its wings in the given time.

For example, if the insect were to fly for 5 seconds, the equation would work as follows:

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1000 = Y

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Step-by-step explanation: hopes this helps

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