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DerKrebs [107]
3 years ago
13

Please help me with this question!! I also need the steps

Mathematics
1 answer:
e-lub [12.9K]3 years ago
8 0

Hi there!

\large\boxed{x = \frac{\pi}{3} \text{ and } \pi}

We are given:

cos(7x)cos(4x) = -1 - sin(7x)sin(4x)

Begin by moving all terms with variables to one side:

cos(7x)cos(4x) + sin(7x)sin(4x) = -1

The corresponding trig identity is cos(A - B). Thus:

cos(7x - 4x) = cos(7x)cos(4x) + sin(7x)sin(4x) = -1

cos(3x) = -1

cos = -1 at  π, so:

3x = π

x = π/3

We can also find another solution. Let 3π = -1:

3x = 3π

x = π

Thus, solutions on [0, 2π) are π/3 and π.

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Two regular hexagons have sides in the ratio 3:5. The area of the smaller hexagon is 81 m2. What is the area of the larger hexag
Aliun [14]
If the sides are in the ratio of 5:3 then the area will increase by a factor of:(5 / 3) ^2
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4 0
3 years ago
Please solve this and only answer with image
horsena [70]

\cot^4 A - \cot^2 A = 1\\\\\cot^4 A = 1 + \cot^2 A\\\\\frac{\cos^4 A}{\sin^4 A} = 1 + \frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A}{\sin^2 A}+\frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A+\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{1}{\sin^2 A}\\\\\cos^4 A = \frac{\sin^4 A}{\sin^2 A}\\\\\cos^4 A = \sin^2 A\\\\

This then means,

\cos^4 A + \cos^2 A = 1\\\\\sin^2 A + \cos^2 A = 1\\\\

which is the pythagorean trig identity. This concludes the proof.

Therefore, if \cot^4 A - \cot^2 A = 1, then \cos^4 A + \cos^2 A = 1

5 0
2 years ago
Please help I don’t understand
Olegator [25]

Answer:

They are not similar.

Step-by-step explanation:

To find whether or not they are similar, set them up as a proportion.

(2.4/5.2) = (8/13)

Now cross multiply.

2.4*13 = 5.2*8

31.2 = 41.6

Since they are not exact, they are not similar.

8 0
3 years ago
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