How do i answer this? which are the bases i should put in the formula?

Mathematics

1 answer:

Nutka1998 [239]4 years ago

5 0

Answer:

The bases are 6 and 3

Step-by-step explanation:

The bases of a trapezoid are the sides that are parallel to each other. The > marks on the lines tells you that those two lines are parallel

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What is the solution of -8/2y-8=5/y+4 - 7y+8/y^2-16

blsea [12.9K]

Answer:

Step-by-step explanation:

$Domain:\\\\2y-8\neq0\ \wedge\ y+4\neq0\ \wedge\ y^2-16\neq0\\\\2y\neq8\ \wedge\ y\neq-4\ \wedge\ y^2\neq16\\\\y\neq4\ \wedge\ y\neq-4\ \wedge\ y\neq\pm\sqrt{16}\\\\y\neq4\ \wedge\ y\neq-4\ \wedge\ y\neq-4\ \wedge\ y\neq4\\\\\boxed{y\neq-4\ \wedge\ y\neq4}\\\\===========================$

$\dfrac{-8}{2y-8}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-16}\\\\\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-4^2}\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{(y-4)(y+4)}\qquad\text{multiply both sides by (-2)}\\\\\dfrac{8}{y-4}=-\dfrac{10}{y+4}+\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{add}\ \dfrac{10}{y+4}\ \text{to both sides}\\\\\dfrac{8}{y-4}+\dfrac{10}{y+4}=\dfrac{14y+16}{(y-4)(y+4)}$

$\dfrac{8(y+4)}{(y-4)(y+4)}+\dfrac{10(y-4)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\\\\\dfrac{8(y+4)+10(y-4)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{use the distributive property}\\\\\dfrac{8y+32+10y-40}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{combine like terms}\\\\\dfrac{(8y+10y)+(32-40)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\\\\\dfrac{18y-8}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\iff18y-8=14y+16\\\\18y-8=14y+16\qquad\text{subtract 14y from both sides}$