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Svetach [21]
3 years ago
12

Eric is swimming along the surface of the sea. He uses positive numbers to represent elevations above the surface of the sea and

negative numbers to represent depths under the surface of the sea. Mark is sitting above Eric and Ariel is diving below Eric. Who could be at an elevation of -3m?
Mathematics
1 answer:
Elza [17]3 years ago
5 0

Answer:

Ariel will be at an elevation of -3 meters.

Step-by-step explanation:

We have been given that Eric is swimming along the surface of the sea. He uses positive numbers to represent elevations above the surface of the sea and negative numbers to represent depths under the surface of the sea.

Since Eric is swimming along the surface of the sea, so his elevation will be 0.

As we are told that Mark is sitting above Eric and Eric uses positive number for elevation above surface of the sea, so Mark's elevation will be positive.

As we are told that Ariel is diving below Eric and Eric uses negative number for elevation below surface of the sea, so Ariel's elevation will be negative.

Therefore, Ariel will be at an elevation of -3 meters.

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When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F
aleksandr82 [10.1K]

Let x be the unknown number. So, three times that number means 3x, and the square of the number is x^2

We have to sum 528 and three times the number, so we have 528+3x

Then, we have to subtract this number from x^2, so we have

x^2-(3x+528)

The result is 120, so the equation is

x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0

This is a quadratic equation, i.e. an equation like ax^2+bx+c=0. These equation can be solved - assuming they have a solution - with the following formula

x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

If you plug the values from your equation, you have

x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}

So, the two solutions would be

x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27

x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24

But we know that x is positive, so we only accept the solution x = 27

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3 years ago
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We will use the sine and cosine of the sum of two angles, the sine and consine of \frac{\pi}{2}, and the relation of the tangent with the sine and cosine:

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\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta

\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0

\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}

If you use those identities, for \alpha=x,\ \beta=\dfrac{\pi}{2}, you get:

\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x

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Hence:

\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x
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Data:
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