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LenaWriter [7]
3 years ago
13

Explain the steps used to solve x - 2 = v(x + 10) and check for extraneous solutions.

Mathematics
1 answer:
Ede4ka [16]3 years ago
8 0
---------------------------------------------------
Question
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x - 2 = √<span>(x + 10) 

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Square both sides
</span>---------------------------------------------------
(x - 2)² = (√(x + 10))²
(x - 2)² = (x + 10)   ← square and square root would will net off each other

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Remove brackets
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x² - 4x + 2² = x + 10
x² - 4x + 4 = x + 10

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minus x + 10 on both sides
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x² - 4x + 4 - x - 10 = 0
x² - 5x - 6 = 0

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Factorise the left side
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(x - 6)(x + 1) = 0

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Apply zero product property 
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x -6 = 0  or  x +1 = 0

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Solve x
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x = 6 or x = -1

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Answer: x = 6 or x = -1
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Consider the following theorem. Theorem If f is integrable on [a, b], then b a f(x) dx = lim n→[infinity] n i = 1 f(xi)Δx where
mel-nik [20]

Split up the interval [1, 9] into <em>n</em> subintervals of equal length (9 - 1)/<em>n</em> = 8/<em>n</em> :

[1, 1 + 8/<em>n</em>], [1 + 8/<em>n</em>, 1 + 16/<em>n</em>], [1 + 16/<em>n</em>, 1 + 24/<em>n</em>], …, [1 + 8 (<em>n</em> - 1)/<em>n</em>, 9]

It should be clear that the left endpoint of each subinterval make up an arithmetic sequence, so that the <em>i</em>-th subinterval has left endpoint

1 + 8/<em>n</em> (<em>i</em> - 1)

Then we approximate the definite integral by the sum of the areas of <em>n</em> rectangles with length 8/<em>n</em> and height f(x_i) :

\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx \approx \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right)

Take the limit as <em>n</em> approaches infinity and the approximation becomes exact. So we have

\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx = \lim_{n\to\infty} \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right) \\\\ = \lim_{n\to\infty} \frac8n \sum_{i=1}^n \left(1+\frac{16}n(i-1)+\frac{64}{n^2}(i-1)^2-4-\frac{32}n(i-1)+6\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=1}^n \left(64(i-1)^2-16n(i-1)+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=0}^{n-1} \left(64i^2-16ni+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(64\sum_{i=0}^{n-1}i^2 - 16n\sum_{i=0}^{n-1}i + 3n^2\sum{i=0}^{n-1}1\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{64(2n-1)n(n-1)}{6} - \frac{16n^2(n-1)}{2} + 3n^3\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{49n^3}3-24n^2+\frac{32n}3\right) \\\\= \lim_{n\to\infty} \frac{8\left(49n^2-72n+32\right)}{3n^2} = \boxed{\frac{392}3}

3 0
3 years ago
I am sooooo dumb i swear,
Citrus2011 [14]

Hei dere!

So this is how I learned it:

18         x

----  =  -------

60       100

Cross cancel it:

60x=1800

x=30%

Hope I helped!

~Potato.

Copyright Potato 2019.

5 0
3 years ago
Kelsie sold digital cameras on her web site. She bought the cameras for $65 each and included a 60% markup to get the selling pr
AysviL [449]

65$*1.60 = 104 dollars...

5 0
3 years ago
Read 2 more answers
Expand 5⁴ WHAT IS THE ANSWER​
Nuetrik [128]

Answer:

5x5x5x5=625

:)

5 0
3 years ago
Help me please ❤️<br> Thank you
anyanavicka [17]

Answer:

I think it would be finding the quantity in each group.

Step-by-step explanation:

Because there would be more than 1 group. there is only one group. so you are trying to find the number of hours each person in the group will get.

i think that is right sorry if not.

5 0
3 years ago
Read 2 more answers
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