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Montano1993 [528]
4 years ago
11

Factorise:7m^2+67m+36

Mathematics
1 answer:
Marta_Voda [28]4 years ago
3 0
You know the first terms of the factors must be 7 & 1, since those are the only factors of 7.

So you write: (7m+x)(m+y)

Both the second and third term are positive, so x & y are both positive.
 
From there, all you can really do is plug in all factors of 36 as x or y until you find that the second term adds up to 67.

And you find the only one that works is:

(7m+4)(m+9)
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Jim's work evaluating 2 (three-fifths) cubed is shown below. 2 (three-fifths) cubed = 2 (StartFraction 3 cubed Over 5 EndFractio
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Answer:

Jim's error is " He did not multiply Three-fifths by 2 before applying the power "

Step-by-step explanation:

Jim's evaluating expression is 2(\frac{3}{5})^3

To verify Jim's error :

Jim's steps are

2(\frac{3}{5})^3

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Therefore 2(\frac{3}{5})^3=\frac{54}{5}

Jim's error is " He did not multiply Three-fifths by 2 before applying the power "

That is the corrected steps are

2(\frac{3}{5})^3

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=2(\frac{27}{125})

=\frac{54}{125}

2(\frac{3}{5})^3=\frac{54}{125}

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4 years ago
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