Bear Valley already had 4 centimeters of snowcover when a snowstorm moved in. The storm

If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn

mash [69]

Let $a=\tan A$ and $b=\sin A$ . Then

$m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab$

$\implies m^2-n^2=4\tan A\sin A$

and

$mn=(a+b)(a-b)=a^2-b^2$

$\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}$

The expression under the square root can be rewritten as

$\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)$

Recall that

$\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A$

so that

$\tan^2A-\sin^2A=\sin^2A\tan^2A$

and assuming $\sin A>0$ and $\tan A>0$ , we end up with

$4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A$

so that

$m^2-n^2=4\sqrt{mn}$

as required.

5 0

3 years ago

Will mark brainliest if answer is relevant and answers first

love history [14]

Step-by-step explanation:

The new price of the flash drive is not $10 because 20% of $10 is 2 and if 2 is marked down by 10% it would not be $10.

6 0

4 years ago

it is known that the population proton of utha residnet that are members of the church of jesus christ 0l6 suppose a random samp

Lady_Fox [76]

Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$