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Kruka [31]
3 years ago
7

ANSWER THIS FAST BRANLIEST IS ON THE LINE in

Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

Option A.) is correct  \triangle XYZ \sim \triangle WVZ  by AA similarity. There are two angles that are congruent given by \angle ZXY = \angle ZWV \hspace{0.2cm}  and\hspace{0.2cm}   \angle ZYX = \angle ZVW.

This is because both pairs represent alternate interior angles formed by lines intersecting two parallel lines.

Step-by-step explanation:

From the given figure we can say that

i) Option A.) is correct  \triangle XYZ \sim \triangle WVZ  by AA similarity. There are two angles that are congruent given by \angle ZXY = \angle ZWV \hspace{0.2cm}  and\hspace{0.2cm}   \angle ZYX = \angle ZVW.

This is because both pairs represent alternate interior angles formed by lines intersecting two parallel lines.

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The maximum value of 12 sin 0-9 sin²0 is: -​
jeka57 [31]

Answer:

4

Step-by-step explanation:

The question is not clear. You have indicated the original function as 12sin(0) - 9sin²(0)

If so, the solution is trivial. At 0, sin(0) is 0 so the solution is 0

However, I will assume you meant the angle to be \theta rather than 0 which makes sense and proceed accordingly

We can find the maximum or minimum of any function by finding the first derivate and setting it equal to 0

The original function is

f(\theta) = 12sin(\theta) - 9 sin^2(\theta)

Taking the first derivative of this with respect to \theta and setting it equal to 0 lets us solve for the maximum (or minimum) value

The first derivative of f(\theta) w.r.t \theta is

                        12\cos\left(\theta\right)-18\cos\left(\theta\right)\sin\left(\theta\right)

And setting this = 0 gives

12\cos\left(\theta\right)-18\cos\left(\theta\right)\sin\left(\theta\right) = 0

Eliminating cos(\theta) on both sides and solving for sin(\theta) gives us

sin(\theta) = \frac{12}{18} = \frac{2}{3}

Plugging this value of sin(\theta) into the original equation gives us

12(\frac{2}{3}) - 9(\frac{4}{9} ) = 8 - 4 = 4

This is the maximum value that the function can acquire. The attached graph shows this as correct

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