Question

Determine as medidas x e y indicadas na figura

Answer 1

Resposta:

3√41²=10+x)²+12²

9.41=10²+2.10.x+x²+144

369=100+20x+x²+144

100+20x+x²+144=369

x²+20x+100+144-369=0

x²+20x-125=0

Δ=20²-4.1.(-125)

Δ=400+500

Δ=900

x1=-20+30/2=10/2=5

x2=-20-30/2=-50/2=-25

x=5

y²=5²+12²

y²=25+144

y²=169

y=√169

y=13

Explicação passo-a-passo: