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Neporo4naja [7]
3 years ago
11

M=6(p+q) solve for q

Mathematics
1 answer:
Katarina [22]3 years ago
3 0
Ok, so m=6(p+q), and solve for q, so it will be like this:
 m=6(p+q)
 m=6p+6q
-6p -6p
 m-6p=6q
 *1/6  *1/6
1/6m-p=q, or you can write it as q=1/6m-p.
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3 years ago
Half a number plus 5 is 11.What is the number?​
SOVA2 [1]
<h3>Solution :</h3>

<u>Let x be the number</u>

\small\bold\red{→}\small\bold{(\frac{1}{2} )x + 5 = 11}

\small\bold\red{→}\small\bold{(\frac{1}{2})x + 5 - 5 = 11 - 5}

\small\bold\red{→}\small\bold{(\frac{1}{2})x = 6}

\small\bold\red{→}\small\bold{2 × (\frac{1}{2})x = 6 × 2 }

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6 0
3 years ago
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RideAnS [48]
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How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not 12?
siniylev [52]

Count the number of multiples of 3, 4, and 12 in the range 1-2005:

⌊2005/3⌋ ≈ ⌊668.333⌋ = 668

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⌊2005/12⌋ ≈ ⌊167.083⌋ = 167

(⌊<em>x</em>⌋ means the "floor" of <em>x</em>, i.e. the largest integer smaller than <em>x</em>, so ⌊<em>a</em>/<em>b</em>⌋ is what you get when you divide <em>a</em> by <em>b</em> and ignore the remainder)

Then using the inclusion/exclusion principle, there are

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numbers that are multiples of 3 or 4 but not 12. We subtract the number multiples of 12 twice because the sets of multiples of 3 and 4 both contain multiples of 12. Subtracting once removes the multiples of 3 <em>and</em> 4 that occur twice. Subtracting again removes them altogether.

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3 years ago
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Answer:

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Step-by-step explanation:

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Combine like terms

x^2 + 11x + 30

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3 years ago
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