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3 years ago

Which ordered pair (x, y) is a solution to the following system of equations?

Mathematics

2 answers:

Rudik [331]3 years ago

7 0

Answer: (-6, 4)

Step-by-step explanation:

You can use the Elimination method:

- Multiply the the first equation by -3 and the second one by 5.

- Add both equations.

- Solve for y:

$\left \{ {{(-3)(5x+4y=-14(-3)} \atop {5(3x+6y)=6(5)}} \right.\\\\\left \{ {{-15x-12y=42} \atop {15x+30y=30}} \right.\\-------\\18y=72\\y=4$

- Susbtittute y=4 into any of the original equations and solve for x:

$3x+6(4)=6\\3x=6-24\\3x=-18\\x=-6$

Then the ordered pair is:

(-6, 4)

Irina18 [472]3 years ago

4 0

<u>Answer:</u>

(-6, 4)

<u>Step-by-step explanation:</u>

We are given the following two equations and we are to solve them:

$5x+4y=-14$ --- (1)

$3x+6y=6$ --- (2)

Using the substitution method:

From equation (2):

$3 x = 6 - 6 y \\\\ x = \frac { 6 - 6 y } { 3 } \\ \\ x = 2 - 2 y$

Substituting this value of x in equation (1) to get:

$5 ( 2 - 2 y ) + 4 y = -14 \\\\ 10 - 10 y + 4 y = -14 \\\\ 1 0 + 14 = 6 y \\\\ y = \frac { 24 } { 6 } \\ \\ y = 4$

Putting this value of y in equation (2) to find the value of x:

$3 x + 6 ( 4 ) = 6 \\\\ 3x + 24 = 6 \\\\ 3x = 6 - 24 \\\\ x = \frac { -18 } { 3 } \\\\ x = -6$

Therefore, (-6, 4) is the solution to the given system of equations.

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3 years ago

4. For each equation, determine if =-2 is a solution. Explain or show your reasoning.

Aneli [31]

Answer:

a. 14 b. -2

Step-by-step explanation:

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3 years ago

The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a m

Mazyrski [523]

Answer:

The probability that the instrument does not fail in an 8-hour shift is $P(X=0) \approx 0.8659$

The probability of at least 1 failure in a 24-hour day is $P(X\geq 1 )\approx 0.3508$

Step-by-step explanation:

The probability distribution of a Poisson random variable X representing the number of successes occurring in a given time interval or a specified region of space is given by the formula:

$P(X)=\frac{e^{-\mu}\mu^x}{x!}$

Let X be the number of failures of a testing instrument.

We know that the mean $\mu = 0.018$ failures per hour.

(a) To find the probability that the instrument does not fail in an 8-hour shift, you need to:

For an 8-hour shift, the mean is $\mu=8\cdot 0.018=0.144$

$P(X=0)=\frac{e^{-0.144}0.144^0}{0!}\\\\P(X=0) \approx 0.8659$

(b) To find the probability of at least 1 failure in a 24-hour day, you need to:

For a 24-hour day, the mean is $\mu=24\cdot 0.018=0.432$

$P(X\geq 1 )=1-P(X=0)\\\\P(X\geq 1 )=1-\frac{e^{-0.432}0.432^0}{0!}\\\\P(X\geq 1 )\approx 0.3508$

3 0

3 years ago

Simplify. (Assume all variables represent positive real numbers). Leave answer in radical form.

Flauer [41]

Answer:

$\sqrt{128a^{6}b^{13}} = 8 a^{3} b^{6} \sqrt{2b}$

Step-by-step explanation:

Given

$\sqrt{128a^{6}b^{13}}$

Required

Solve

$\sqrt{128a^{6}b^{13}}$

The expression can be split to:

$\sqrt{128a^{6}b^{13}} = \sqrt{128} * \sqrt{a^{6}} * \sqrt{b^{13}}$

$\sqrt{128a^{6}b^{13}} = \sqrt{64 * 2} * \sqrt{a^{6}} * \sqrt{b^{13}}$

$\sqrt{128a^{6}b^{13}} = \sqrt{64} * \sqrt{2} * \sqrt{a^{6}} * \sqrt{b^{13}}$

$\sqrt{128a^{6}b^{13}} = \sqrt{64} * \sqrt{2} * \sqrt{a^{6}} * \sqrt{b^{12 + 1}}$

$\sqrt{128a^{6}b^{13}} = \sqrt{64} * \sqrt{2} * \sqrt{a^{6}} * \sqrt{b^{12}} * \sqrt{b}$

So, we have:

$\sqrt{128a^{6}b^{13}} = 8 * \sqrt{2} * a^{6/2} * b^{12/2} * \sqrt{b}$

$\sqrt{128a^{6}b^{13}} = 8 * \sqrt{2} * a^{3} * b^{6} * \sqrt{b}$

Rewrite as:

$\sqrt{128a^{6}b^{13}} = 8 * a^{3} * b^{6}* \sqrt{2} * \sqrt{b}$

$\sqrt{128a^{6}b^{13}} = 8 a^{3} b^{6}* \sqrt{2b}$

$\sqrt{128a^{6}b^{13}} = 8 a^{3} b^{6} \sqrt{2b}$

5 0

2 years ago