Answer:
i think probably 80°
Step-by-step explanation:
PQ = RS
2x+3 = 4x-3
x = 3
PTQ = RTS
9y-3x-1 = 5x+7y-5
2y-8x = -4
y-4x = -2
y-4(3) = -2
y-12 = -2
y = -2+12
y = 10
so PTQ = 9y-3x-1
9(10)-3(3)-1
90-9-1
80°
8/15 = 56/105
This may be solved using proportion.
a/b = c/d
ad = bc where cross products are equal.
a = 8
b = f
c = 56
d = 105
ad = bc
8*105 = f*56
840 = 56f
840/56 = 56f/56
f = 15
56 ÷ 7 = 8
105 ÷ 7 = 15
34.06 is the the correct way of saying it.
Answer:
z (min) = 705
x₁ = 10
x₂ = 9
Step-by-step explanation:
Let´s call x₁ quantity of food I ( in ou ) and x₂ quantity of food II ( in ou)
units of vit. C units of vit.E Cholesterol by ou
x₁ 32 9 48
x₂ 16 18 25
Objective function z
z = 48*x₁ + 25*x₂ To minimize
Subject to:
1.-Total units of vit. C at least 464
32*x₁ + 16*x₂ ≥ 464
2.- Total units of vit. E at least 252
9*x₁ + 18*x₂ ≥ 252
3.- Quantity of ou per day
x₁ + x₂ ≤ 35
General constraints x₁ ≥ 0 x₂ ≥ 0
Using the on-line simplex method solver (AtoZmaths) and after three iterations the solution is:
z (min) = 705
x₁ = 10
x₂ = 9
