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3 years ago

Solve the following System of Equations 5x+4y-z=1 2x-2y+z=1 -x-y+z=2

Mathematics

1 answer:

podryga [215]3 years ago

3 0

Answer:

x = 0 , y = 1 , z = 3

Step-by-step explanation:

Solve the following system:

{5 x + 4 y - z = 1 | (equation 1)

2 x - 2 y + z = 1 | (equation 2)

-x - y + z = 2 | (equation 3)

Subtract 2/5 × (equation 1) from equation 2:

{5 x + 4 y - z = 1 | (equation 1)

0 x - (18 y)/5 + (7 z)/5 = 3/5 | (equation 2)

-x - y + z = 2 | (equation 3)

Multiply equation 2 by 5:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

-x - y + z = 2 | (equation 3)

Add 1/5 × (equation 1) to equation 3:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x - y/5 + (4 z)/5 = 11/5 | (equation 3)

Multiply equation 3 by 5:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x - y + 4 z = 11 | (equation 3)

Subtract 1/18 × (equation 2) from equation 3:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x+0 y+(65 z)/18 = 65/6 | (equation 3)

Multiply equation 3 by 18/65:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Subtract 7 × (equation 3) from equation 2:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y+0 z = -18 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Divide equation 2 by -18:

{5 x + 4 y - z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Subtract 4 × (equation 2) from equation 1:

{5 x + 0 y - z = -3 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Add equation 3 to equation 1:

{5 x+0 y+0 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Divide equation 1 by 5:

{x+0 y+0 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Collect results:

Answer: {x = 0 , y = 1 , z = 3

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Answer:

Option D) Yes, because the test statistic is -2.01

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30 pound

Sample mean, $\bar{x}$ = 29.1 pounds

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = 2 pounds

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We use one-tailed(left) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

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Answer:

$f'(x) = 4x^2 - 3$ for $x \le -3$

Step-by-step explanation:

See attachment for proper question

Given

$f(x) = -\frac{1}{2}\sqrt{x + 3}$

For

$x \ge -3$

Required

Determine the inverse function

$f(x) = -\frac{1}{2}\sqrt{x + 3}$

Replace f(x) with y

$y = -\frac{1}{2}\sqrt{x + 3}$

Swap the positions of x and y

$x = -\frac{1}{2}\sqrt{y + 3}$

Multiply both sides by -2

$-2 * x =-2 * -\frac{1}{2}\sqrt{y + 3}$

$-2x =\sqrt{y + 3}$

Square both sides

$(-2x)^2 =(\sqrt{y + 3})^2$

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Make y the subject

$y = 4x^2 - 3$

The inverse has been solved. So, we need to replace y with f'(x)

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Next, is to determine the interval

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Change inequality to $\le$

$x \le -3$

Hence, the inverse function is:

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