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Vladimir79 [104]
3 years ago
12

Segment AB falls on line 6x + 3y = 12. Segment CD falls on line 4x + 2y = 8. What is true about segments AB and CD?

Mathematics
2 answers:
astra-53 [7]3 years ago
8 0

Answer:

C. They are lines that lie exactly on top of one another because they have the same slope and the same y-intercep

Step-by-step explanation:

spin [16.1K]3 years ago
6 0

Answer:

  • C. They are lines that lie exactly on top of one another because they have the same slope and the same y-intercept.

Step-by-step explanation:

<u>Given lines</u>

  • 6x + 3y = 12 ⇒ 2x + y = 4 ⇒ y = -2x + 4
  • 4x + 2y = 8 ⇒ 2x + y = 4 ⇒ y = -2x + 4

The lines are same as both have same slope and y-intercept.

<u>Correct statement is following:</u>

  • C. They are lines that lie exactly on top of one another because they have the same slope and the same y-intercept.
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y= 3x -5

Step-by-step explanation:

since when x is 0, and y is -5, you know that the y intercept is -5.

look at the table

1x -5 = -2

x= 3

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B) 8xy +4y² factorise
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The answer is: 4y(2x+y)

Step-by-step explanation:

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The entrance to our school has a height of 10.5 feet. There are
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Step-by-step explanation:

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3 years ago
The length of a rectangle solid is 3.0 m the width is 0.6 meter, and the height is 0.4 meter. Find, to the nearest tenth, the nu
Ad libitum [116K]

Answer:

0.72\,\,m^3

Step-by-step explanation:

Given:

Length of a rectangle solid = 3 m

Width of a rectangle solid = 0.6 m

Height of a rectangle solid = 0.4 m

To find: Volume of the solid

Solution:

Volume of the solid = length × breadth × height

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So, the number of cubic meters in the volume of the solid is 0.72\,\,m^3

3 0
3 years ago
Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

6 0
2 years ago
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