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2 years ago

Help me with this equation Please

Mathematics

1 answer:

kykrilka [37]2 years ago

3 0

Answer:

$2598.80

Step-by-step explanation:

49 at $12.80 per ticket = $627.20

106 at $18.60 per ticket = $1971.60

Total = $2598.80

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Please please help!!

shepuryov [24]

Answer:

$\large\boxed{x=0\ and\ x=\pi}$

Step-by-step explanation:

$\tan^2x\sec^2x+2\sec^2x-\tan^2x=2\\\\\text{Use}\ \tan x=\dfrac{\sin x}{\cos x},\ \sec x=\dfrac{1}{\cos x}:\\\\\left(\dfrac{\sin x}{\cos x}\right)^2\left(\dfrac{1}{\cos x}\right)^2+2\left(\dfrac{1}{\cos x}\right)^2-\left(\dfrac{\sin x}{\cos x}\right)^2=2\\\\\left(\dfrac{\sin^2x}{\cos^2x}\right)\left(\dfrac{1}{\cos^2x}\right)+\dfrac{2}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=2$

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

$\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{(1+\cos^2x)(\cos^2x)}{(\cos^2x)^2}=2\qquad\text{Use the distributive property}\\\\\dfrac{1-\cos^2x+\cos^2x+\cos^4x}{\cos^4x}=2\\\\\dfrac{1+\cos^4x}{\cos^4x}=2\qquad\text{multiply both sides by}\ \cos^4x\neq0\\\\1+\cos^4x=2\cos^4x\qquad\text{subtract}\ \cos^4x\ \text{from both sides}\\\\1=\cos^4x\iff \cos x=\pm\sqrt1\to\cos x=\pm1\\\\ x=k\pi\ for\ k\in\mathbb{Z}\\\\\text{On the interval}\ 0\leq x$

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3 years ago

I NEED HELP NOW I WILL DO ANYTHING

frosja888 [35]

The answer is C if its not tell me but if so give me 5 stars and send me a friend request.

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3 years ago

SQUARE ROOTS<br> 2x2 +50 =-166

Nataly_w [17]

Answer:

Its a wrong i teaching 2*2=4 Cause 4+50=54 -Its a correct

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3 years ago

Which graph shows a linear function?

Stolb23 [73]

Answer:

Step-by-step explanation:

You have shared only one graph, that of a quadratic function with vertex at (0, 0) and an equation based upon y = ax^2, where a is a constant coefficient.

Please go back to the source of your question and describe the other graphs that were given to you.

The graph that shows a straight line is the linear function.

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3 years ago

A sports news station wanted to know whether people who live in the North or the South are bigger sports fans. For its study, 17

Free_Kalibri [48]

Using the z-distribution, it is found that the 95% confidence interval for the difference is (-1.3, -0.7).

<h3>What are the mean and the standard error for each sample?</h3>

Considering the data given:

$\mu_S = 3.8, n = 175, s_S = \frac{1.7}{\sqrt{175}} = 0.1285$

$\mu_N = 4.8, n = 152, s_N = \frac{1.2}{\sqrt{152}} = 0.0973$

<h3>What is the mean and the standard error for the distribution of differences?</h3>

The mean is the subtraction of the means, hence:

$\overline{x} = \mu_S - \mu_N = 3.8 - 4.8 = -1$

The standard error is the square root of the sum of the variances of each sample, hence:

$s = \sqrt{s_S^2 + s_N^2} = \sqrt{0.1285^2 + 0.0973^2} = 0.1612$

<h3>What is the confidence interval?</h3>

It is given by:

$\overline{x} \pm zs$

We have a 95% confidence interval, hence the critical value is of z = 1.96.

Then, the bounds of the interval are given as follows:

$\overline{x} - zs = -1 - 1.96(0.1612) = -1.3$

$\overline{x} + zs = -1 + 1.96(0.1612) = -0.7$

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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2 years ago