3 years ago

How do u solve 3y-4=6-2y3y-4=6-2y

Mathematics

1 answer:

Travka [436]3 years ago

3 0

Answer:

y=2

Step-by-step explanation:

3y+2y-4=6

3y+2y=6+4

5y=10

5y over 5=10 divide 5

y=2

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Plssss help quick i need it for sum important

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(a) The product of $(2x-4)\ and\ (3x^2-x+4)$ is calculated to be $=6x^3-14x^2+12x-16$

(b)The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

<u>Step-by-step explanation:</u>

$(2x-4)(3x^2-x+4)\\\\=(2x.3x^2)+(2x.-x)+(2x.4)+(-4.3x^2)+(-4.-x)+(-4.4)\\\\=6x^3-2x^2+8x-12x^2+4x-16\\\\=6x^3-14x^2+12x-16$

$(4x-2)(3x^2-x+4)\\\\=(4x.3x^2)+(4x.-x)+(4x.4)+(-2.3x^2)+(-2.-x)+(-2.4)\\\\=12x^3-4x^2+16x-6x^2+2x-8\\\\=12x^3-10x^2+18x-8$

The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

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