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goldenfox [79]
4 years ago
14

Solve the equation 6w2 – 7w – 20 = 0.

Mathematics
1 answer:
Irina-Kira [14]4 years ago
4 0

Answer:

\large\boxed{B.\ w=-\dfrac{4}{3},\ w=\dfrac{5}{2}}

Step-by-step explanation:

6w^2-7w-20=0\\\\6w^2-15w+8w-20=0\\\\3w(2w-5)+4(2w-5)=0\\\\(2w-5)(3w+4)=0\iff2w-5=0\ \vee\ 3w+4=0\\\\2w-5=0\qquad\text{add 5 to both sides}\\2w=5\qquad\text{divide both sides by 2}\\\boxed{w=\dfrac{5}{2}}\\\\3w+4=0\qquad\text{subtract 4 from both sides}\\3w=-4\qquad\text{divide both sides by 3}\\\boxed{w=-\dfrac{4}{3}}

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Please let be know if I have interpreted your question correctly. Thank you.

Step-by-step explanation:

Part a)

This is the way I interpret part a:

Find integers a,b,c such that a|(b-c) but a does not divide b and a does not divide c.

There are many triples satisfying part a.

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3 divides 10-7

But 3 does not divide 10 and 3 does not divide 7.

Another example is (a,b,c)=(2,7,5).

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But 2 doesn't divide 7 and 2 doesn't divide 5.

I will give another example.

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But 10 doesn't divide 19 and 10 doesn't divide 9

Part b)

This is the say I interpret part b:

a | (2b + 3c) but a does not divide b and a does not divide c

So lets let b=2 and c=5, then 2b+3c=19.

We just need to choose an a such that it divides 19 but not 2 and 5. That should be easy. 19 itself will do that.

So one (a,b,c) could be (19,2,5).

Let's try for another example.

Let b=3 and c=1.

Then 2b+3c=9.

So we just need to find an a that divides 9 but not 3 and not 1.

9 works.

So another example is (9, 3,1).

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