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adell [148]
2 years ago
14

2(x + 2) + 2 = 2(x + 3) + 1

Mathematics
1 answer:
Maslowich2 years ago
6 0

Answer:

No solution

Step-by-step explanation:

5(x + 4) – x = 4(x + 5) – 1

x = ?

Solution:

5(x+4)−x=4(x+5)−1

Apply Distributive property:

  • (5)(x)+(5)(4)−x=(4)(x)+(4)(5)−1
  • 5x+20−x=4x+20−1

Combine Like Terms:

  • (5x−x)+(20)=(4x)+(20−1)
  • 4x+20=4x+19
  • 4x+20=4x+19

Subtract 4x from both sides:

  • 4x+20−4x=4x+19−4x
  • 20=19

Subtract 20 from both sides:

  • 20−20=19−20
  • 0=−1

There isn't any solutions.

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soldi70 [24.7K]

Answer:

24×36×5=4,320

4,320÷6=720

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How many x intercepts appear on the graph of this polynomial function? f(x)=x^4-5x^2 1 x intercept 2 x intercepts 3 x intercepts
Bumek [7]
You can find the number of x intercepts by setting f(x) equal to zero giving you 0=x^4-5x^2. You can factor out an x^2 and divide on both sides (when you divide 0 by anything it is still zero so the x^2 disappears), and you are left with 0=x^2-5, which you can solve easily giving you x=+/- sqrt(5). However, the original equation also has an x intercept at (0,0) which you can see by plugging 0 in for x. So the grand total of x intercepts is 3!
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lois put 8 liters of water in a bucket for her pony.at the end of the day, there were 2 liters of water left.how much water did
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If lim x-&gt; infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b
MAXImum [283]

We have

\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}

So

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0

and this happens if <em>a</em> = 1 and <em>b</em> = -1.

To confirm, we have

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0

as required.

3 0
2 years ago
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