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3 years ago

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: SimplifyingQx + 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,

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