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Gekata [30.6K]
3 years ago
9

P(x) = x + 1x² – 34x + 343 d(x)= x + 9

Mathematics
1 answer:
Feliz [49]3 years ago
5 0

Answer:

x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > \frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1. Go ahead and simplify.

x=\frac{9}{d-1};\quad \:d\ne \:1.

Now, \mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}.

P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343.

Group the like terms... 1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343.

\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1} > 1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343.

Now for 1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}.

\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}.

Now for 33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}.

Thus we then get \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343.

Now we want to combine fractions. \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}.

\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2

\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}

\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}.

Expand 81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac.

-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1.

\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}

Therefore P=\frac{-297d+378}{\left(d-1\right)^2}+343.

Hope this helps!

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A chemist has one solution that has 50% acid. She has another solution that is 25% acid. How many liters of the 50% acid solutio
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Answer:

Step-by-step explanation:

Let's begin by assigning letters to represent our two unknowns:

   x (liters of 25% solution)

   y (liters of 50% solution)

 

Our system of equations will consist of two equations:

   Equation #1 (total volume of solution)

   Equation #2 (total concentration of acid)

 

Our total volume of solution is 10 liters, which can be expressed as the sum of our unknowns:

   Equation #1:  x + y = 10

 

Our total concentration of acid can be expressed as the sum of the individual acid concentrations to make up the concentration of the final solution:

   Equation #2:  (0.25)(x) + (0.50)(y) =

                            (0.40)(10)

 

We can use Equation #1 to express one unknown in terms of the other and then plug that expression into Equation #2 to solve for one of the unknowns:

   x + y = 10

         y = 10 - x

 

Now we'll plug our expression for y in terms of x into Equation #2 and solve for x:

   0.25(x) + 0.50(10 - x) = 0.40(10)

   0.25x + 5 - 0.50x = 4

   -0.25x = 4 - 5

   -0.25x = -1

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          x = 4 (liters of 25% solution)

 

Now we'll plug our value for x into Equation #1 and solve for y:

   4 + y = 10

         y  = 10 - 4

         y = 6 (liters of 50% solution)

 

Finally, we will verify the correctness of our answers by plugging these values into Equation #2 to see if the sum of the component acid concentrations equals the final solution concentration:

   0.25(4) + (0.50)(6) = 0.40(10)

   1 + 3 = 4

         4 = 4 (our answers are correct)

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2 years ago
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