1) n(6;1;-1)=n(A;B;C), point A(2;9;4)=A(x₀;y₀;z₀);
2) common view for a plane according the conditions is: A(x-x₀)+B(y-y₀)+C(z-z₀)=0;
3) after substituted coordinates: 6(x-2)+(y-9)-(z-4)=0; ⇒ 6x+y-z-17=0.
If the weight of the bottom box is b then the weight of a stack of four boxes is b + b/2 + b/4 + b/8
= 8b/8 + 4b/8 + 2b/8 + 1b/8
= 15/8 b
A stack must weigh less than 100 pounds
15/8 b < 100
b < 100 x 8/15 = 53 1/3 pounds
If boxes have a minimum weight of 1 pound then
for stacks that contain 4 boxes the range of b is [8, 53 1/3)
and if stacks contain 1 to 4 boxes the range of b is [1, 53 1/3)
If the only restriction is a maximum of 100 then the range of b is [0, 53 1/3)
Answer: A = X + Y; B = X xor Y. The task is to make X as minimum as possible.
Step-by-step explanation:
i think the graph g(x) has been shrunk vertically by a factor of 1/5 and shifted up by 4 units :)