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Blizzard [7]
3 years ago
12

Solve the following system of equations x + y = 3 x + 2y = 6

Mathematics
1 answer:
lyudmila [28]3 years ago
7 0

In this question, you're solving the systems of equations to find what x and y equal to.

Solve the systems of equations:

x + y = 3

x + 2y = 6

Lets make the "y" variable the same by multiplying the top equation by -2.

-2(x + y = 3)

x + 2y = 6

----------------

-2x -2y = -6

x + 2y = 6

Solve:

-x =0

x = 0

Now, we know x = 0, now plug 0 into x in one of the equations and solve:

We'll plug it into the x + y = 3 equation.

0 + y = 3

y = 3

When you're done solving, you should get:

x = 0 and y = 3

Answer:

x = 0

y = 3

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g The downtime per day for a computing facility has mean 4 hours and standard deviation 0.9 hour. What assumptions must be true
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To obtain a valid approximation for probabilities about the average daily downtime, either the underlying distribution(of the downtime per day for a computing facility) must be normal, or the sample size must be of 30 or more.

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In this question:

To obtain a valid approximation for probabilities about the average daily downtime, either the underlying distribution(of the downtime per day for a computing facility) must be normal, or the sample size must be of 30 or more.

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Parallel lines cut by transversal?..help?
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2 years ago
Find all solutions of each equation on the interval 0 ≤ x < 2π.
Korvikt [17]

Answer:

x = 0 or x = \pi.

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}.

\displaystyle \sec{x} = \frac{1}{\cos{x}}.

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, \sin^{2}{x} = 1 - \cos^{2}{x}. Therefore, for the square of tangents,

\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}.

This equation will thus become:

\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2.

To simplify the calculations, replace all \cos^{2}{x} with another variable. For example, let u = \cos^{2}{x}. Keep in mind that 0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1.

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Solve this equation for u:

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Given that 0 \le x < 2\pi,

0 \le k.

k = 0 or k = 1. Accordingly,

x = 0 or x = \pi.

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3 years ago
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