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katrin [286]
3 years ago
11

Qx+12=13x+P with only one solution

Mathematics
2 answers:
ra1l [238]3 years ago
4 0
What are you solving for? if you are solving for x, then x=(P-12)/(Q-13)
AlexFokin [52]3 years ago
3 0
The answer to the question is: Simplifying<span>Qx + 12 = 13x + P Reorder the terms: 12 + xQ = 13x + P Reorder the terms: 12 + xQ = P + 13x Solving 12 + xQ = P + 13x Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-13x' to each side of the equation. 12 + -13x + xQ = P + 13x + -13x Combine like terms: 13x + -13x = 0 12 + -13x + xQ = P + 0 12 + -13x + xQ = P Add '-12' to each side of the equation. 12 + -13x + -12 + xQ = -12 + P Reorder the terms: 12 + -12 + -13x + xQ = -12 + P Combine like terms: 12 + -12 = 0 0 + -13x + xQ = -12 + P -13x + xQ = -12 + P Reorder the terms: 12 + -1P + -13x + xQ = -12 + P + 12 + -1P Reorder the terms: 12 + -1P + -13x + xQ = -12 + 12 + P + -1P Combine like terms: -12 + 12 = 0 12 + -1P + -13x + xQ = 0 + P + -1P 12 + -1P + -13x + xQ = P + -1P Combine like terms: P + -1P = 0 12 + -1P + -13x + xQ = 0 The solution to this equation could not be determined. Hope this helps:)

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The annual inventory cost C for a manufacturer is given below, where Q is the order size when the inventory is replenished. Find
shtirl [24]

Answer:

Change in annual cost is 1.63 (decreasing).

Instantaneous rate of change is 1.65 (decreasing).

Step-by-step explanation:

Given that,

C=\dfrac{1017000}{Q}+6.8\:Q

The annual change in cost is given by,  

Change\:in\:C=C\left(348\right)-C\left(347\right)

Calculate the value of cost at Q = 347 and Q =348 by substituting the value in cost function.

Calculating value of C at Q=347,

C=\dfrac{1017000}{347}+6.8\left(347\right)

C=5290.44

Calculating value of C at Q=348,

C=\dfrac{1017000}{348}+6.8\left(348\right)}

C= 5288.81

Substituting the value,  

Change\:in\:C=5288.81-5290.44

Change\:in\:C=-1.63

Negative sign indicate that there is decrease in annual cost when Q is increased from 347 to 348

Therefore, change is annual cost is 1.63  (decreasing).

Instantaneous rate of change is given by the formula,  

f’\left(x\right)=\lim_{h\rightarrow 0}\dfrac{f\left(x+h\right)-f\left(x\right)}{h}

Rewriting,

C’\left(Q\right)=\lim_{h\rightarrow 0}\dfrac{C\left(Q+h\right)-C\left(Q\right)}{h}

Now calculate \dfrac{C\left(Q+h\right) by substituting the value Q+h in cost function,

C\left(Q+h\right)= \dfrac{1017000}{Q+h}+6.8\left(Q+h\right)

Therefore,  

C’\left(Q\right)= \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}+6.8\left(Q+h\right)-\left(\dfrac{1017000}{Q}+6.8Q\right)}{h}

By using distributive law,

C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}+6.8Q+6.8h-\dfrac{1017000}{Q}-6.8Q}{h}

Cancelling out common factors,  

C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}-\dfrac{1017000}{Q}+6.8h}{h}

Now, LCD of \left(Q+h\right) and Q is Q(Q+h). So multiplying first term by Q  and second term by

Therefore,  

C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}\times\dfrac{Q}{Q}-\dfrac{1017000}{Q}\times\dfrac{Q+h}{Q+h}+6.8h}{h}

Simplifying,  

C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q}{\left(Q+h\right)\left(Q\right)}-\dfrac{1017000\left(Q+h\right)}{Q\left(Q+h\right)}+6.8h}{h}

C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q-1017000\left(Q+h\right)}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}

C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q-1017000Q-1017000h}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}

C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{-1017000h}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}

Factoring out h from numerator,  

C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{h\left(\dfrac{-1017000}{\left(Q+h\right)\left(Q\right)}+6.8\right)}{h}

Cancelling out h,  

C'\left(Q\right) = \lim_{h\to 0}\: -\dfrac{1017000}{\left(Q+h\right)Q}+6.8

Calculating the limit by plugging value h = 0,

C'\left(Q\right) = -\dfrac{1017000}{\left(Q+0\right)Q}+6.8

C'\left(Q\right) = -\dfrac{1017000}{Q^{2}}+6.8

Given that Q=347,

C'\left(347\right) = -\dfrac{1017000}{347^{2}}+6.8

C'\left(347\right) = -1.65

Negative sign indicate that there is decrease in instantaneous rate when Q is 347

Therefore, instantaneous rate of change at Q=347 is 1.65  (decreasing).

4 0
3 years ago
Jessel poured 3/4 of a carton of milk into his glass. He drank 2/3 of what he took. What part of the carton did he drink?​
lesya [120]

Answer:

Since he poured 3/4 of milk into the glass and you have to find 2/3 of it you multiply both fractions gettin 6/12 you can reduce that to 1/2

6 0
3 years ago
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mixer [17]
She makes 3 dozen, or 36 an hour
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A board game has a spinner divided into sections of equal size. Each section is labeled between 1 and 5
RideAnS [48]
What is the actual question?
3 0
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2. Find f(-1).<br> f(x) = x2-4x+2<br> X+7<br> Round your answer to the nearest hundredth.
natka813 [3]

Step-by-step explanation:

f(x)=\dfrac{x^2-4x+2}{x+7}\\\\f(-1)=\text{put}\ x=-1\ \text{to}\ f(x):\\\\f(-1)=\dfrac{(-1)^2-4(-1)+2}{-1+7}=\dfrac{1+4+2}{6}=\dfrac{7}{6}

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