If 3(d+1)=4(d–2),
then what is the value of 5(d–2)?

9514 1404 393

  45

Step-by-step explanation:

Solving for d, we find ...

  3(d +1) = 4(d -2)

  3d +3 = 4d -8 . . . . . . eliminate parentheses

  11 = d  . . . . . . . . . . . .  subtract 3d-8

Then the value of 5(d-2) is ...

  5(11 -2) = 5(9) = 45

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The graphical solution gives both the value of "d" and the value of 4(d-2). To get 5(d-2), we only need to add d-2 to the value of 4(d-2).

  5(d-2) = (11 -2) +36 = 45

6 equilateral triangles

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: