-8y^4*(?)=32y^11
11 answers
-8y^4 * (-4y^7) = 32y^11
(?)(7y^5-3y)=35y^14-15y10
(5y^9 - 3y^5) = 35y^14 - 15y^10
(?)(7y^5-3y)=35y^14-15y10
what is the value of the (?)
what is the value of the (?)
The value of the ? is 5y^9.
This is because 5y^9 multiplied by (7y^5 - 3y) results in 35y^14 - 15y^10:
5y^9 * 7y^5 = 35y^14
5y^9 * -3y = -15y^10
This is because 5y^9 multiplied by (7y^5 - 3y) results in 35y^14 - 15y^10:
5y^9 * 7y^5 = 35y^14
5y^9 * -3y = -15y^10
-7y(3y^6+2y)
-21y^7 - 14y^2
5y^4(5y^7+8y^3-9y^2)
25y^11 + 40y^7 - 45y^6
(3y^4+5) (5y^7+4)
15y^11 + 20y^7 + 25y^4 + 20