3 years ago

Find the real or imaginary solutions of the equation by factoring. 6x^2+13x-5=0

1 answer:

4 0

$6x^2+13x-5=0\\
6x^2-2x+15x-5=0\\
2x(3x-1)+5(3x-1)=0\\
(2x+5)(3x-1)=0\\
x=-\dfrac{5}{2} \vee x=\dfrac{1}{3}$

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kupik [55]

To factor it'd be (3x-2) (3x-2)

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3 years ago

grandymaker [24]

Step-by-step explanation:

The statement in the above question is True.

Sum of three prime numbers (other than two) is always odd.

Going by Christian Goldbach number theory ,

Goldbach stated that every odd whole number greater than 5 can be written as sum of three prime numbers .

Lets take an example,

Later on in 2013 the Mathematician Harald Helfgott proved this theory true for all odd numbers greater than five.

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3 years ago

PolarNik [594]

Answer:

$8>3$

Step-by-step explanation:

STEP 1: Rewrite so n is on the left side of the inequality.

$6n+8-2n>4n+3$

STEP 2: Subtract 2 n from 6 n .

$4n+8>4n+3$

STEP 3: Move all terms containing n to the left side of the inequality.

$8>3$

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3 years ago

gogolik [260]

The correct answer is below:

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3 years ago

statuscvo [17]

6 would be correct answer

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3 years ago