90 yards= 3/4 let d be the distance. (3/4)d=90 Then solve by dividing both sides by d.
The trend will be positive because as the students get older their heights tend to get taller.
This is all a bit complicated so try and stick with me on this one!
This one is the first problem in the first picture! a + 2 / a^2 + a - 1 / a + 1/ -a^2 - 2a - 1
= a^4 + a^3 - a^2 - a / -a^5 - 3a^4 - 3a^3 - a^2
= -a^3 - a^2 + a + 1 / a^4 + 3a^3 + 3a^2 + a
= -a^3 - a^2 + a + 1/ a^4 + 3a^3 + 3a^2 + a
= (-a - 1) (a + 1) (a - 1) / a(a+ 1) (a + 1) (a + 1)
= -a + 1 / a^2 + a
The second problem in the first picture! 3x / y + 3x - y^2 / 3xy - 9x^2 + y^2 + 9x^2 / y^2 - 9x^2
= -81x^4y + 18x^2y^3 - y^5 / 243x^5 - 54x^3y^2 + 3xy^4
This one is for the last picture! 4y^2 + 4y + 1 / 4y - 8y^2 - 4y^2 + 1 / 4y + y
= 16y^3 + 24y^2 / -32y^3 + 16u^2
= 16y + 24 / -32y + 16
= 2y + 3 / -4y + 2
I hope this was helpful!!!
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