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2 years ago

What is 2 to the 5th power multiplied by 2 to the negative 3rd power??

Mathematics

1 answer:

e-lub [12.9K]2 years ago

8 0

Answer:

Step-by-step explanation:

$2^5 \times 2^{-3}$

Because the powers have the same base of 2, we can apply the index law:

$a^b \times a^c=a^{b+c}$

Therefore, $2^5 \times 2^{-3}=2^{5-3}=2^2=4$

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Find cot and cos <br> If sec = -3 and sin 0 > 0

Natali5045456 [20]

Answer:

Second answer

Step-by-step explanation:

We are given $\displaystyle \large{\sec \theta = -3}$ and $\displaystyle \large{\sin \theta > 0}$ . What we have to find are $\displaystyle \large{\cot \theta}$ and $\displaystyle \large{\cos \theta}$ .

First, convert $\displaystyle \large{\sec \theta}$ to $\displaystyle \large{\frac{1}{\cos \theta}}$ via trigonometric identity. That gives us a new equation in form of $\displaystyle \large{\cos \theta}$ :

$\displaystyle \large{\frac{1}{\cos \theta} = -3}$

Multiply $\displaystyle \large{\cos \theta}$ both sides to get rid of the denominator.

$\displaystyle \large{\frac{1}{\cos \theta} \cdot \cos \theta = -3 \cos \theta}\\\displaystyle \large{1=-3 \cos \theta}$

Then divide both sides by -3 to get $\displaystyle \large{\cos \theta}$ .

Hence, $\displaystyle \large{\boxed{\cos \theta = - \frac{1}{3}}}$

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Next, to find $\displaystyle \large{\cot \theta}$ , convert it to $\displaystyle \large{\frac{1}{\tan \theta}}$ via trigonometric identity. Then we have to convert $\displaystyle \large{\tan \theta}$ to $\displaystyle \large{\frac{\sin \theta}{\cos \theta}}$ via another trigonometric identity. That gives us:

$\displaystyle \large{\frac{1}{\frac{\sin \theta}{\cos \theta}}}\\\displaystyle \large{\frac{\cos \theta}{\sin \theta}$

It seems that we do not know what $\displaystyle \large{\sin \theta}$ is but we can find it by using the identity $\displaystyle \large{\sin \theta = \sqrt{1-\cos ^2 \theta}}$ for $\displaystyle \large{\sin \theta > 0}$ .

From $\displaystyle \large{\cos \theta = -\frac{1}{3}}$ then $\displaystyle \large{\cos ^2 \theta = \frac{1}{9}}$ .

Therefore:

$\displaystyle \large{\sin \theta=\sqrt{1-\frac{1}{9}}}\\\displaystyle \large{\sin \theta = \sqrt{\frac{9}{9}-\frac{1}{9}}}\\\displaystyle \large{\sin \theta = \sqrt{\frac{8}{9}}}$

Then use the surd property to evaluate the square root.

Hence, $\displaystyle \large{\boxed{\sin \theta=\frac{2\sqrt{2}}{3}}}$

Now that we know what $\displaystyle \large{\sin \theta}$ is. We can evaluate $\displaystyle \large{\frac{\cos \theta}{\sin \theta}}$ which is another form or identity of $\displaystyle \large{\cot \theta}$ .

From the boxed values of $\displaystyle \large{\cos \theta}$ and $\displaystyle \large{\sin \theta}$ :-

$\displaystyle \large{\cot \theta = \frac{\cos \theta}{\sin \theta}}\\\displaystyle \large{\cot \theta = \frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}}}\\\displaystyle \large{\cot \theta=-\frac{1}{3} \cdot \frac{3}{2\sqrt{2}}}\\\displaystyle \large{\cot \theta=-\frac{1}{2\sqrt{2}}$

Then rationalize the value by multiplying both numerator and denominator with the denominator.

$\displaystyle \large{\cot \theta = -\frac{1 \cdot 2\sqrt{2}}{2\sqrt{2} \cdot 2\sqrt{2}}}\\\displaystyle \large{\cot \theta = -\frac{2\sqrt{2}}{8}}\\\displaystyle \large{\cot \theta = -\frac{\sqrt{2}}{4}}$

Hence, $\displaystyle \large{\boxed{\cot \theta = -\frac{\sqrt{2}}{4}}}$

Therefore, the second choice is the answer.

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Summary

Trigonometric Identity

$\displaystyle \large{\sec \theta = \frac{1}{\cos \theta}}\\ \displaystyle \large{\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}}\\ \displaystyle \large{\sin \theta = \sqrt{1-\cos ^2 \theta} \ \ \ (\sin \theta > 0)}\\ \displaystyle \large{\tan \theta = \frac{\sin \theta}{\cos \theta}}$