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3 years ago

Please help asap! Will give brainliest! Please answer correctly!

Mathematics

2 answers:

MAXImum [283]3 years ago

8 0

Answer:

6^21

Step-by-step explanation:

We know that a^b^c = a^(b*c)

6^7^3 = 6^(7*3) = 6^21

shtirl [24]3 years ago

3 0

Answer:

Option D

Step-by-step explanation:

Rule : $(a^{b})^{c} = a^{b.c}$

For example : $(5^{3})^{4} = 5^{3.4}= 5^{12}$

Your question: $(6^{7})^{3} = 6^{7.3}= 6^{21}$

Hope this helps ^-^

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spayn [35]

Answer:

4, 9v, and 6w

Step-by-step explanation:

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3 years ago

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Mice21 [21]

Power and chain rule (where the power rule kicks in because $\sqrt x=x^{1/2}$ ):

$\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'$

Simplify the leading term as

$\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}$

Quotient rule:

$\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}$

Chain rule:

$(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)$

$(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$

Put everything together and simplify:

$\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}$

$=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}$

$=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}$

5 0

3 years ago

Help what is 2v+5(3v+4

Artemon [7]

explanation:

There isn't an exact way to answer this problem. Is there more to the question? If not, I am unable to answer it for you.

4 0

3 years ago

What is the slope and y intercept of 6x + 24 =0

yuradex [85]

X = -4 is your answer

6 0

3 years ago

If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that the d

katen-ka-za [31]

Answer:

We need a sample size of least 119

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Sample size needed

At least n, in which n is found when $M = 0.09$

We don't know the proportion, so we use $\pi = 0.5$ , which is when we would need the largest sample size.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.09 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.09\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.09}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.09})^{2}$

$n = 118.6$

Rounding up

We need a sample size of least 119

6 0

3 years ago