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2 years ago

Solve this equation, and show work. worth 40 points.

Mathematics

1 answer:

ioda2 years ago

7 0

Answer:

$r=-3/5=-0.6$

Step-by-step explanation:

We have the equation:

$6^{-3r}=6^{2r+3}$

Since their bases are the same, the exponents must be equivalent. So:

$-3r=2r+3$

Solve for r, subtract 2r from both sides:

$-5r=3$

Divide both sides by -5:

$r=-3/5=-0.6$

So, the value of r is -3/5.

And we're done!

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In ΔOPQ, the measure of ∠Q=90°, the measure of ∠O=26°, and QO = 4.9 feet. Find the length of PQ to the nearest tenth of a foot.

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Given:

In ΔOPQ, m∠Q=90°, m∠O=26°, and QO = 4.9 feet.

To find:

The measure of side PQ.

Solution:

In ΔOPQ,

$m\angle O+m\angle P+m\angle Q=180^\circ$ [Angle sum property]

$26^\circ+m\angle P+90^\circ=180^\circ$

$m\angle P+116^\circ=180^\circ$

$m\angle P=180^\circ -116^\circ$

$m\angle P=64^\circ$

According to Law of Sines, we get

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

Using the Law of Sines, we get

$\dfrac{p}{\sin P}=\dfrac{o}{\sin O}$

$\dfrac{QO}{\sin P}=\dfrac{PQ}{\sin O}$

Substituting the given values, we get

$\dfrac{4.9}{\sin (64^\circ)}=\dfrac{PQ}{\sin (26^\circ)}$

$\dfrac{4.9}{0.89879}=\dfrac{PQ}{0.43837}$

$\dfrac{4.9}{0.89879}\times 0.43837=PQ$

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