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grigory [225]
3 years ago
6

Substitute x+2y=8 3x+2y=6

Mathematics
2 answers:
vekshin13 years ago
8 0
Use x+2y=8
Subtract 2y to make
x=8-2y

Plug this in to the other equation
3(8-2y)+2y=6

Distribute the 3
24-6y+2y=6

Combine like terms
24-4y=6

Solve
18/4=y, 9/2=y

Plug and solve for x
X+2y=8
X+(2*9/2)=8
X+9=8
X=-1
Zigmanuir [339]3 years ago
6 0
X + 2y = 8

3x+2y = 6

x + 2y =8
   -2y     -2y

x = 8 - 2y

Now replace the value for x into the second equation. 

3(8-2y) - 2y

Distribute property

3*8 = 24
3*2y = 6y

24 - 6y - 2y = 6

24 - 8y = 6
-24           -24 

-8y = -18

-8y/8 = -18/8

y = 2.25

Now replace the value for y into the equation.

x + 2(2.25) = 8
x + 4.5 = 8
    -4.5     -4.5

x = 3.5

Now we have our two values.

x = 3.5
y = 2.25
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Encuentre el radio y el área de círculos cuya circunferencias tienen las siguientes medidas
Vadim26 [7]

Answer:

Los radios de los círculos son r_{1} \approx 9.995, r_{2} \approx 2.013, r_{3} \approx 7.592, respectivamente.

Las áreas de los círculos son A_{1} \approx 313.845, A_{2} \approx 12.730, A_{3} \approx 181.077, respectivamente.

Step-by-step explanation:

La circunferencia (s) se calcula mediante la siguiente fórmula:

s = 2\pi\cdot r (1)

Donde r es el radio del círculo.

Una vez hallado el radio, se determina el área de la figura geométrica (A) mediante la siguiente fórmula:

A = \pi\cdot r^{2} (2)

Si conocemos que las circunferencias son s_{1} = 62.8, s_{2} = 12.65 y s_{3} = 47.7, respectivamente:

1) Radios de los círculos

r_{1} = \frac{s_{1}}{2\pi}, r_{2} = \frac{s_{2}}{2\pi}, r_{3} = \frac{s_{3}}{2\pi}

r_{1} \approx 9.995, r_{2} \approx 2.013, r_{3} \approx 7.592

2) Áreas de los círculos

A_{1} = \pi\cdot r_{1}^{2}, A_{2} = \pi\cdot r_{2}^{2}, A_{3} = \pi\cdot r_{3}^{2}

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Two fair dice are rolled. Determine whether the events are dependent or independent. (a) (the first die shows 3) and (the two di
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Answer: The events are independent

a) p = 1/36 b) p = 5/216

Step-by-step explanation: both events are independent because the fact that the first die shows it sample space {1, 2, 3, 4, 5, 6} does not stop the second die from showing it own sample space too which is {1, 2, 3, 4, 5, 6}.

The sample space of first die = {1, 2, 3, 4, 5, 6}

The sample space of second die = {1, 2, 3, 4, 5, 6}

The sample space of 2 dice rolled at once is seen as attachment to this answer, find attachment below.

Probability that an event will occur = number of time the event will occur / total number of possible outcome of the event

a)

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