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

(4y-3)(2y²+3y-5)

First , let's start with "4y"

4y*2y² = 8y³

4y*3y =  12y²

4y*-5 = -20y

Next, let's multiply by "-3"

-3*2y² = -6y²

-3*3y = -9y

-3*-5 = 15

Now, let's combine all of our values.

8y³+12y²-6y²-20y-9y+15 = 8y³+6y²-29y+15

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

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