Respuesta :
Answer:
(A)[tex]A=\dfrac12h(b_1+b_2)[/tex]
(D)[tex]b_1=\dfrac{2(A-\dfrac12 hb_2)}{h}[/tex]
Step-by-step explanation:
Given that:
[tex]A=\dfrac12h(b_1+b_2)[/tex]
[tex]2A=h(b_1+b_2)\\2A=hb_1+hb_2\\hb_1=2A-hb_2\\b_1=\dfrac{2A-hb_2}{h} \\b_1=\dfrac{2(A-\dfrac12 hb_2)}{h}[/tex]
Therefore, A and D are equivalent
