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MArishka [77]
2 years ago
6

Solve for x: 5/2 - x = 5x - 2x

Mathematics
1 answer:
Dmitrij [34]2 years ago
8 0

Step-by-step explanation:

5/2 - x = 5x - 2x

5/2 - x = 3x

Bringing like terms on one side

5/2 = 3x + x

5/2 = 4x

2.5 = 4x

2.5/4 = x

0.625 = x

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andrew-mc [135]

- 3x + 4y = 7 \\ y = 3x - 5

  • <em>[</em><em>Substitute</em><em> </em><em>the</em><em> </em><em>given</em><em> </em><em>value</em><em> </em><em>of</em><em> </em><em>y</em><em> </em><em>into</em><em> </em><em>the</em><em> </em><em>first</em><em> </em><em>equation</em><em>]</em>

<em>- 3x + 4 \times (3x - 5) = 7</em>

  • <em>[</em><em>Distribute</em><em> </em><em>4</em><em> </em><em>through</em><em> </em><em>the</em><em> </em><em>parentheses</em><em>]</em>

<em>- 3x + 12x - 20 = 7</em>

  • <em>[</em><em>Collect</em><em> </em><em>like</em><em> </em><em>terms</em><em>]</em>

<em>9x - 20 = 7</em>

  • <em>[</em><em>Move</em><em> </em><em>the</em><em> </em><em>constant</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>right-hand</em><em> </em><em>side</em><em> </em><em>and</em><em> </em><em>change</em><em> </em><em>its</em><em> </em><em>sign</em><em>]</em>

<em>9x = 7 + 20</em>

  • <em>[</em><em>Add</em><em> </em><em>the</em><em> </em><em>numbers</em><em>]</em>

<em>9x = 27</em>

  • <em>[</em><em>Divide</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em> </em><em>by</em><em> </em><em>9</em><em>]</em>

<em>x = 3</em>

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<em>y = 3 \times 3 - 5</em>

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<em>y = 4</em>

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5 5/6 ÷ 3 3/4 - 2/9=
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Answer:

\boxed{ \: ans = \frac{5}{9} }

Step-by-step explanation:

5 5/6 ÷ 3 3/4 - 2/9 \to \:  \frac{35}{6}  \div  \frac{15}{4}  -  \frac{2}{9}  \\  \frac{35}{6}  \times  \frac{4}{15}  -  \frac{2}{9}  \\  \frac{7}{9}  -  \frac{2}{9}  \\   \boxed{ \: ans = \frac{5}{9} }

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Euler's method:
Leokris [45]

Answer:

f(1) ≈ 2.7864

Step-by-step explanation:

You appear to want a couple of iterations of ...

... y[n+1] = y[n] +arcsin(x[n]·y[n]}·(x[n+1] -x[n])

... x[n+1] = x[n] +0.5

... x[0] = 0

... y[0] = 2

Filling in the values, we get

... y[1] = 2 + arcsin(0·2)·0.5 = 2

... y[2] = 2 + arcsin(0.5·2)·0.5 = 2 +(π/2)·0.5 ≈ 2.7864 . . . . corresponds to x=1

7 0
3 years ago
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