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Harlamova29_29 [7]
1 year ago
12

Meldie made a cube whose side is (3x - 2) inches. Find its volume in cubic inches.

Mathematics
1 answer:
irinina [24]1 year ago
5 0

Answer:

x^2(3x-2) cubic inches OR in^3

OR

3x { 3 [ 3x ( x - 2 ) + 4 ] } - 8 cubic inches OR in^3

I AM UNAWARE IF YOU ASKED THAT ONE SIDE IS (3X-2) OR ALL. I WILL ANSWER BOTH PARTS

<em>-</em>

<em>NOTE</em><em>:</em><em> </em><em>'</em><em>^</em><em>'</em><em> </em><em>MEANS</em><em> </em><em>TO</em><em> </em><em>THE</em><em> </em><em>POWER</em><em> </em><em>OF</em><em>.</em><em>.</em>

<em>-</em>

Volume = v, abc = 3 sides of cube (height, width, length)

Using the formula for volume in a cube,

v = abc

We can solve this.

If one side is (3x-2)in,

  • (3x-2)(x)(x) = v.... x are the other two sides
  • x^2(3x-2) = v

x^2(3x-2) cubic inches OR in^3

If all sides are (3x-2)in,

Use the formula,

(a - b) ^{3}  =  {a}^{3}  + 3ab(b - a) -  {b}^{3}

We can solve this.

  • (3x-2)(3x-2)(3x-2) = v
  • (3x-2)^3 = v.... 3x = a and -2 = b
  • (3x)^3 + [(3)(3x)(2)][2-3x] - (2)^3 = v
  • 27x^3 + 18x(2-3x) -8 = v
  • (27x^3 + 36x - 54x^2) - 8 = v.. Terms inside brackets - take 3x as common and leave out 8
  • 3x(9x^2 -18x +12) = v... Take 3 as common again in the brackets
  • 3x [ 3 ([3x^2 -6x] + 4) -8 = v....Take 3x common in the terms in square brackets
  • 3x [ 3 [ 3x (x-2) + 4 ]] - 8 = v
  • 3x { 3 [ 3x ( x - 2 ) + 4 ] } - 8 = v

3x { 3 [ 3x ( x - 2 ) + 4 ] } - 8 cubic inches OR in^3

___

If you have any questions regarding formulas or anything, comment and I will get back to you asap.

___

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x^3y^2+\sin(x\ln y)+e^{xy}=0

Differentiate both sides, treating y as a function of x. Let's take it one term at a time.

Power, product and chain rules:

\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}

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Product and chain rules:

\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}

=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)

=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)

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Product and chain rules:

\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}

=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)

=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)

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The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0

Isolate the derivative, and solve for it:

\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by y to get rid of that fraction in the denominator.

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}

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Step-by-step explanation:

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Step-by-step explanation:

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