Mathematics: Find the values of y: 3y^3(y^2+ 2/3y)=0

Find the values of y: 3y^3(y^2+ 2/3y)=0 - id33223011894, lowealiyah6755

Ответ:

Ravish6667

09.03.2022

Mathematics

y = 0 or y = -2/3

Step-by-step explanation:

$3y^3\left(y^2+\dfrac{2}{3}y\right)=0$

The product is zero if one of the factors is equal to zero. Therefore

$3y^3\left(y^2+\dfrac{2}{3}y\right)=0\iff3y^3=0\ \vee\ y^2+\dfrac{2}{3}y=0\\\\3y^3=0\qquad\text{divide both sides by 3}\\\\y^3=0\to \boxed{y=0}\\\\y^2+\dfrac{2}{3}y=0\\\\y\left(y+\dfrac{2}{3}\right)=0\iff y=0\ \vee\ y+\dfrac{2}{3}=0\to \boxed{y=-\dfrac{2}{3}}$

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Ответ:

arroyoe342

09.03.2022

Mathematics

y =0 y = -2/3

Step-by-step explanation:

3y^3(y^2+ 2/3y)=0

Factor out a y

3y^3 *y (y+2/3) =0

3y^4 (y+2/3) =0

Using the zero product property

3y^4 =0 y+2/3 =0

y^4 = 0 y+2/3-2/3 =0

y =0 y = -2/3

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Ответ:

Anliz0179

16.04.2023

Mathematics

You should set up an X and Y table. Make time X and Distance Y. (the table is like a cross: on the left top put x and left top put y. Then under the line list the x and y numbers given in order.) Then find the "slope" which is y/x (rise over run) then you get the increase rate of 45 mph. The relation between them would be D=45t

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Find the values of y: 3y^3(y^2+ 2/3y)=0

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