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4 years ago

5

Aimee is flying a plane for the first time! The air traffic controller tells her to fly at 500m above the ground or higher. Writ

e an inequality that is true only for elevations (h) at which Aimee should fly.

1 answer:

Gelneren [198K]4 years ago

6 0

She needs to fly at 500 or higher which means that the equation needs to include
$\geqslant$

The answer is
$h \geqslant 500$

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Simplify tan9x-tan5x / 1+tan9xtan5x<br> (SHOW WORK)

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Answer:

The simplest form is tan(4x)

Step-by-step explanation:

* Lets revise the identity of the compound angles

- $tan(a+b)=\frac{tan(a)+tan(b)}{1-tan(a)tan(b)}$

- $tan(a-b)=\frac{tan(a)-tan(b)}{1+tan(a)tan(b)}$

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- Let 9x = 5x + 4x

∴ tan(9x) = tan(5x + 4x)

- Use the rule of the compound angle

∵ $\frac{tan(9x)-tan(5x)}{1+tan(9x)tan(5x)}$ ⇒ (1)

∵ $tan(5x+4x)=\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)}$ ⇒ (2)

∵ tan(9x) = equation (2)

- Substitute (2) in (1)

∴ $\frac{\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)}-tan(5x)}{1+(\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)})tan(5x)}$

- Multiply up and down by (1 - tan(5x)tan(4x))

∴ $\frac{tan(5x)+tan(4x)-tan(5x)[1-tan(5x)tan(4x)]}{1-tan(5x)tan(4x)+tan(5x)[tan(5x)+tan(4x)]}$

- Simplify up and down

∴ $\frac{tan(5x)+tan(4x)-tan(5x)+tan^{2}(5x)tan(4x)}{1-tan(5x)tan(4x)+tan^{2}(5x)+tan(5x)tan(4x) }$

∴ $\frac{tan(4x)+tan^{2}(5x)tan(4x)}{[1+tan^{2}(5x)]}$

- Take tan(4x) as a common factor up

∴ $\frac{tan(4x)[1+tan^{2}(5x)]}{[1+tan^{2}(5x)]}$

- Cancel [1 + tan²(5x)] up and down

∴ The answer is tan(4x)

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