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matrenka [14]
3 years ago
9

Solve the equation

Mathematics
1 answer:
Ksivusya [100]3 years ago
4 0

Answer:

x = -2/5

Step-by-step explanation:

x + 3 - 2x = 4x + 5

Combine like terms

-x + 3 = 4x + 5

Subtract 4x from both sides

-5x + 3 = 5

Subtract 3 from both sides

-5x = 2

Divide both sides by -5

x = -2/5

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7th grade help me plzzzzz
Gekata [30.6K]

Answer:

fifteen, "number sentence"

Step-by-step explanation:

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Convert the following degree measure to radian measure 30 degrees
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Step-by-step explanation:

given 30°

converting into radian measure

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Read 2 more answers
What is -15x=0 solution
cricket20 [7]
Here is the method for solving that equation:

Step #1:  Write the equation:

             -15x = 0

Step #2:  Divide each side of the equation by -15 :

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Five adult tickets and three child tickets for a movie costs £55. The cost of buying two adult tickets and three child tickets i
Allisa [31]

Answer:

  • adult £8
  • child £5

Step-by-step explanation:

If you look at the numbers you are given, you see that the first purchase has 3 more adult tickets than the second purchase, and its cost is £24 more. This means an adult ticket costs £24/3 = £8.

Two adult tickets will cost 2×£8 = £16, so three child tickets cost ...

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Each child ticket is then £15/3 = £5.

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7 0
3 years ago
What are the potential solutions of In(x^2-25)=0?
Maslowich

Answer:

x =\pm \sqrt{26}

Step-by-step explanation:

<u>Given </u><u>:</u><u>-</u><u> </u>

  • ln ( x² - 25 ) = 0

And we need to find the potential solutions of it. The given equation is the logarithm of x² - 25 to the base e . e is Euler's Number here. So it can be written as ,

<u>Equation</u><u> </u><u>:</u><u>-</u><u> </u>

\implies log_e {(x^2-25)}= 0

<u>In </u><u>general</u><u> </u><u>:</u><u>-</u><u> </u>

  • If we have a logarithmic equation as ,

\implies log_a b = c

Then this can be written as ,

\implies a^c = b

In a similar way we can write the given equation as ,

\implies e^0 = x^2 - 25

  • Now also we know that a^0 = 1 Therefore , the equation becomes ,

\implies 1 = x^2 - 25 \\\\\implies x^2 = 25 + 1 \\\\\implies x^2 = 26 \\\\\implies x =\sqrt{26} \\\\\implies x = \pm \sqrt{ 26}

<u>Hence</u><u> the</u><u> </u><u>Solution</u><u> </u><u>of </u><u>the</u><u> given</u><u> equation</u><u> is</u><u> </u><u>±</u><u>√</u><u>2</u><u>6</u><u>.</u>

6 0
3 years ago
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