Question

determine o valor dos elementos desconhecidos

Answer 1

teorema de Pitágoras

a)
${y}^{2} = {3}^{2} + {7}^{2}$
${y}^{2} = 9 + 49$
${y}^{2} = 58$
$y = 7.7...$
b)
$(2 \sqrt{13} ) {}^{2} = {x}^{2} + {6}^{2}$
$4 \times 13 = {x}^{2} + 36$
${x}^{2} = 58 - 36$
${x}^{2} = 22$
$x = 4.6...$
c)
$(y + 6) {}^{2} = {y}^{2} + {8}^{2}$
${y}^{2} + 12y + 36 = {y}^{2} + 64$
$12y = 28$
$y = 2.3...$
d)

$(x + 2) {}^{2} = (x + 1) {}^{2} + {x}^{2}$
${x}^{2} + 4x + 4 = {x}^{2} + 2x + 1 + {x}^{2}$
$- {x}^{2} + 2x + 3 = 0$
$delta = 4 - 4.( - 1).3$
$delta = 4 + 12 = 16$
$x1 = \frac{ - 2 + 4}{ - 2} = - 1$
$x2 = \frac{ - 2 - 4}{ - 2} = 3$

Answer 2

Resposta:

a) y²=3²+7²

    y=√9+49

    y=√58

b) (2√13)²=x²+6²

4*13=x²+36

 52-36=x²

x²=16⇒x=√16=4

c) (y+6)²= y²+8²

y²+12y+36=y²+64

  12y=64-36

   12y=28

       y=28/12=7/3

  d) (x+2)²=x²+(x+1)²

 x²+4x+4=x²+x²+2x+1

   x²+2x+1-4x-4=0

   x²-2x-3=0

   Δ=4+12⇒Δ=16

   x=2±√16/2

   x'=2+4/2

    x'=6/2=3

     x"=2-4/2

     x"=-2/2=-1

Explicação passo-a-passo: