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3 years ago

Find all solutions of the equation in the interval [0°, 360°). cos(x) = -1

Mathematics

1 answer:

nika2105 [10]3 years ago

7 0

Answer:

180°

Step-by-step explanation:

cos(x) = - 1 has only 1 solution in the required interval , that is

x = $cos^{-1}$ (- 1) = 180°

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A medication states that the odds of having an allergic reaction are 1 in 50. What is that probability states as a percent?

Fittoniya [83]

<h2>Answer:</h2>

<u><em>As a percent, the odds of having an allergic reaction would be 2%.</em></u>

This is because 1 in 50 of a chance, is equal to 1/50, and to make this a percent out of 100, just multiply both parts of the fraction by 2.

None of the choices equal 1/50, so let's just make it a percentage anyway.

1/50 ---> 0.02

0.02 ---> 2%

<em>The answer is 2%, or 1/50 as a fraction.</em>

3 0

4 years ago

5(-7)+9(4) I am confused on this problem

amm1812

Answer:

the answer is 1

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

4 years ago

A simple random sample of 60 items resulted in a sample mean of 65. The population standard deviation is 14. a. Compute the 95%

bazaltina [42]

Answer:

The 95% confidence interval for the population mean is between 61.5 and 68.5.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So: