Answer:
z (min) = 150090.8 ( monetary units ) ( $ )
x₁ = 2000 x₄ = 0 x₂ = 3382 x₅ = 368 x₃ = 0 x₆ = 1700
Step-by-step explanation:
wrenches produced in-house ( W111 = x₁ W222 = x₂ W333 = x₃ )
x₁ x₂ and x₃
wrenches produced outside (W111 = x₄ W222 = x₅ W333 = x₆ )
x₄ x₅ and x₆
Objective function:
z = 17*x₁ + 20.40*x₄ + 19*x₂ + 21.85*x₅ + 23*x₃ + 25.76*x₆ to minimize
First constraint: Manufacturing hours: 16500
2.5*x₁ + 3.4*x₂ + 3.8*x₃ ≤ 16500
Second constraint: Inspection hours : 1600
0.25*x₁ + 0.3*x₂ + 0.45*x₃ ≤ 1600
Three demands constraint:
x₁ + x₄ = 2000
x₂ + x₅ = 3750
x₃ + x₆ = 1700
x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 x₄ ≥ 0 x₅ ≥ 0 x₆ ≥ 0 all integers
After 6 iterations with on-line solver the solution is:
z (min) = 150090.8 ( monetary units ) ( $ )
x₁ = 2000 x₄ = 0 x₂ = 3382 x₅ = 368 x₃ = 0 x₆ = 1700
The answer is d because when you go from positive to negative you add the positive number to get to zero and than add the number you are getting to make sense?
Answer:
= 2√6
Step-by-step explanation:
a² + b² = c²
1 + b² = 25
b² = 24
b = ±√24
b = ±2√6
We can forget about the minus answer because you cannot have a negative length. That does not make sense.
Please mark me brainliest.
Given:
20 ounces of water
someone finishes 55% of water
What is left?
20 × 0.55 (55% =

= 0.55)
11
20 - 11 = 9
There are 9 ounces of water left

It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if
</span><span>

</span><span>In notation we write respectively
</span>

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence

Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²<span>
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c</span>²

That is to say, if c = -2, f(x) is continuous at x = 4.
Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers 