What is the product ? (4y-3)(2y^2+3y-5)

Mathematics

2 answers:

DIA [1.3K]3 years ago

5 0

Answer:

=8y^3+6y^2−29y+15

Step-by-step explanation:

(4y−3)(2y^2+3y−5)

=(4y+−3)(2y^2+3y+−5)

=(4y)(2y^2)+(4y)(3y)+(4y)(−5)+(−3)(2y^2)+(−3)(3y)+(−3)(−5)

=8y^3+12y^2−20^y−6y^2−9y+15

=8y^3+6y^2−29y+15

yuradex [85]3 years ago

4 0

Answer:

8y^3 + 6y^2 - 29y + 15

Step-by-step explanation:

8y^3 + 12y^2 - 20y - 6y^2 - 9y + 15

then

8y^3 + 6y^2 - 29y + 15 is the product

Best regards

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The population of Ohio will be 10 million in the year 1970

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$10=\frac{12.79}{1+2.402e^{-}0.0309t}\\\\10(1+2.402e^{-}0.0309t)=12.79\\\\1+2.402e^{-}0.0309t=\frac{12.79}{10} \\\\1+2.402e^{-}0.0309t=1.279\\\\2.402e^{-}0.0309t=1.279-1\\\\2.402e^{-}0.0309t = 0.279\\\\e^{-}0.0309t =\frac{0.279}{2.402} \\\\e^{-}0.0309t =0.116\\\\-0.0309t=ln0.116\\\\-0.0309t=-2.154\\\\t=\frac{-2.154}{-0.0309} \\\\t=69.7$