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aev [14]
3 years ago
10

Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x? StartFraction 5 plus-or-minus 3 StartRoot 11 EndRoo

t Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 47 EndRoot Over 12 EndFraction StartFraction negative 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction
Mathematics
2 answers:
Licemer1 [7]3 years ago
5 0

Answer:

Using the quadratic formula to solve 5x = 6x^2 – 3, what are the values of x?

5x = 6x^2 – 3

Subtract 5x from both sides:

0 = 6x^2 – 5x – 3

a = 6, b = -5, c = -3

x = (-b ± √(b^2 - 4ac))/(2a)

x = (-(-5) ± √((-5)^2 - 4(6)(-3)))/(2(6))

x = (5 ± √(25 + 72))/12

x = (5 ± √(97))/12

Step-by-step explanation:

iris [78.8K]3 years ago
3 0

Answer:

your answer is C

Step-by-step explanation:

thank me later;)

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A box with a square base and open top must have a volume of 296352 c m 3 . We wish to find the dimensions of the box that minimi
mestny [16]

Answer:

  • Base Length of 84cm
  • Height of 42 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 296352 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume, V=x^2h=296352

h=\dfrac{296352}{x^2}

Surface Area of the box = Base Area + Area of 4 sides

A(x,h)=x^2+4xh\\$Substitute h=\dfrac{296352}{x^2}\\A(x)=x^2+4x\left(\dfrac{296352}{x^2}\right)\\A(x)=\dfrac{x^3+1185408}{x}

Step 2: Find the derivative of A(x)

If\:A(x)=\dfrac{x^3+1185408}{x}\\A'(x)=\dfrac{2x^3-1185408}{x^2}

Step 3: Set A'(x)=0 and solve for x

A'(x)=\dfrac{2x^3-1185408}{x^2}=0\\2x^3-1185408=0\\2x^3=1185408\\$Divide both sides by 2\\x^3=592704\\$Take the cube root of both sides\\x=\sqrt[3]{592704}\\x=84

Step 4: Verify that x=84 is a minimum value

We use the second derivative test

A''(x)=\dfrac{2x^3+2370816}{x^3}\\$When x=84$\\A''(x)=6

Since the second derivative is positive at x=84, then it is a minimum point.

Recall:

h=\dfrac{296352}{x^2}=\dfrac{296352}{84^2}=42

Therefore, the dimensions that minimizes the box surface area are:

  • Base Length of 84cm
  • Height of 42 cm.
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