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NeTakaya
3 years ago
15

Please help asap y= ?

Mathematics
1 answer:
Tanzania [10]3 years ago
5 0

Answer:

y = 2|x| - 3

Step-by-step explanation:

We'll start with our parent function which is y = |x|. Notice that graph is translated down by 3 units. So, we'll add that:

y = |x| - 3

Also notice that the ordered pair (1, -1) is on the graph. |1| = 1, so a translation of -3 should take it to -2 and it should have been (1, -2). But, it's not so it must be that the graph is stretched/compressed. If you multiply every point by 2, it gives us the result we want. Say for example, f(1) = 2|1| - 3 = 2 - 3  = -1. We get the point (1, -1) which is exactly what we want. We write compressions as (where a is the compression/stretching constant):

y = a|x| + d

Our compression constant is 2, so our final answer is:

<em>y = 2|x| - 3</em>.

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Answer:

Choice A. 3.

Step-by-step explanation:

The triangle in question is a right triangle.

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As a result, the length of both legs can be found directly using the sine function and the cosine function.

Let \text{Opposite} denotes the length of the side opposite to the 30^{\circ} acute angle, and \text{Adjacent} be the length of the side next to this 30^{\circ} acute angle.

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The longer leg in this case is the one adjacent to the 30^{\circ} acute angle. The answer will be 3.

There's a shortcut to the answer. Notice that \sin{30^{\circ}} < \cos{30^{\circ}}. The cosine of an acute angle is directly related to the adjacent leg. In other words, the leg adjacent to the 30^{\circ} angle will be the longer leg. There will be no need to find the length of the opposite leg.

Does this relationship \sin{\theta} < \cos{\theta} holds for all acute angles? (That is, 0^{\circ} < \theta?) It turns out that:

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