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2 years ago

Ss below. help would be appreciated.

Mathematics

2 answers:

Arada [10]2 years ago

7 0

Answer:

(3,0,0)

Step-by-step explanation:

3x + 5y + z = 9 --------------(I)

-4y - z = 0 -------------------(II)

z = -4y

2x + 3y - 4z = 6 (III)

Substitute z = -4y in equation (I) and (III)

3x +5y - 4y = 9

3x + y = 9 -----------------------(a)

2x + 3y - 4*(-4y) = 6

2x + 3y + 16y = 6

2x + 19y = 6 -------------------(b)

Multiply equation (a) by (-19) and then add. So, y will be eliminated and we can get the value of x

(a)*(-19) -57x - 19y = -171

(b) 2x + 19y = 6

-55x = -165

x = -165/-55

x = 3

Plugin x = 3 in equation (a)

3*3 + y = 9

9 + y = 9

y = 9 - 9

y = 0

Plug in y = 0 in equation (II)

-4*0 - z = 0

z = 0

Step-by-step explanation:

sergeinik [125]2 years ago

5 0

Answer:

(3,0,0)

Step-by-step explanation:

3x + 5y + z = 9 --------------(I)

-4y - z = 0 -------------------(II)

z = -4y

2x + 3y - 4z = 6 (III)

Substitute z = -4y in equation (I) and (III)

3x +5y - 4y = 9

3x + y = 9 -----------------------(a)

2x + 3y - 4*(-4y) = 6

2x + 3y + 16y = 6

2x + 19y = 6 -------------------(b)

Multiply equation (a) by (-19) and then add. So, y will be eliminated and we can get the value of x

(a)*(-19) -57x - 19y = -171

(b) <u> 2x + 19y = 6 </u>

-55x = -165

x = -165/-55

x = 3

Plugin x = 3 in equation (a)

3*3 + y = 9

9 + y = 9

y = 9 - 9

y = 0

Plug in y = 0 in equation (II)

-4*0 - z = 0

z = 0

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When studying radioactive material, a nuclear engineer found that over 365 days,

Gnom [1K]

Answer:

a) The mean number of radioactive atoms that decay per day is 81.485.

b) 0% probability that on a given day, 50 radioactive atoms decayed.

Step-by-step explanation:

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\lambda$ is the mean in the given interval.

a. Find the mean number of radioactive atoms that decayed in a day.

29,742 atoms decayed during 365 days, which means that:

$\lambda = \frac{29742}{365} = 81.485$

The mean number of radioactive atoms that decay per day is 81.485.

b. Find the probability that on a given day, 50 radioactive atoms decayed.

This is P(X = 50). So

$P(X = 50) = \frac{e^{-81.485}*(81.485)^{50}}{(50)!} = 0$

0% probability that on a given day, 50 radioactive atoms decayed.

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2 years ago

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Akimi4 [234]

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3 years ago

Find all the zeros of the polynomial function p(x) = x3 – 5x2 + 33x – 29

hram777 [196]

Answer:

$\large \boxed{\sf \ \ x=1, \ \ x=2+5i, \ \ x=2-5i \ \ }$

Step-by-step explanation:

Hello,

I assume that we are working in $\mathbb{C}$ , otherwise there is only one zero which is 1. Please consider the following.

First of all, <u>we can notice that 1 is a trivial solution</u> as

$p(1) = 1^3-5\cdot 1^2 + 33\cdot 1-29=1-5+33-29=0$

It means that (x-1) is a factor of p(x) so we can find two real numbers, a and b, so that we can write the following.

$p(x)=(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=x^3+(a-1)x^2+(b-a)x-b$

Let's identify like terms as below.

a-1 = -5 <=> a = -5 + 1 = -4

b-a = 33

-b = -29 <=> b = 29

$\boxed{ \ p(x)=(x-1)(x^2-4x+29) \ }$

Now, we need to find the zeroes of the second factor, meaning finding x so that:

$x^2-4x+29=0 \ \text{ complete the square, 29 = 25 + 4} \\ \\ x^2-2\cdot 2 \cdot x+2^2+25=0 \\ \\ (x-2)^2=-25=(5i)^2 \ \text{ take the root } \\ \\x-2=\pm 5i \ \text{ add 2 } \\ \\ x = 2+5i \ \text{ or } \ x = 2-5i$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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