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Elenna [48]
2 years ago
13

Eight more than seven times a number is equal to four more than five times the number. What is the number?

Mathematics
2 answers:
Alex17521 [72]2 years ago
8 0

Answer: -2

Step-by-step explanation:

convert it to a equation:

7x+8=5x+4

move x's to a side and numbers to the other

2x=-4

divide by 2

x=-2

Digiron [165]2 years ago
8 0

Let us consider the number to be 'n'.

<u>Given condition</u> :

  • 7n + 8 = 5n + 4

<u>Subtract '5n' from each side</u> :

  • 7n + 8 - 5n = 5n + 4 - 5n
  • 2n + 8 = 4

<u>Subtract '8' from each side</u> :

  • 2n + 8 - 8 = 4 - 8
  • 2n = -4

<u>Divide '2' from each side</u> :

  • 2n/2 = -4/2
  • \fbox {n = -2}

∴ The number is -2.

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Answer:

1) S = 2\cdot w\cdot l - 8\cdot x^{2}, 2) The domain of S is 0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}. The range of S is 0 \leq S \leq 2\cdot w \cdot l, 3) S = 176\,in^{2}, 4) x \approx 4.528\,in, 5) S = 164.830\,in^{2}

Step-by-step explanation:

1) The function of the box is:

S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)

S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w

S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x

S = 2\cdot w\cdot l - 8\cdot x^{2}

2) The maximum cutout is:

2\cdot w \cdot l - 8\cdot x^{2} = 0

w\cdot l - 4\cdot x^{2} = 0

4\cdot x^{2} = w\cdot l

x = \frac{\sqrt{w\cdot l}}{2}

The domain of S is 0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}. The range of S is 0 \leq S \leq 2\cdot w \cdot l

3) The surface area when a 1'' x 1'' square is cut out is:

S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}

S = 176\,in^{2}

4) The size is found by solving the following second-order polynomial:

20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}

20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}

8\cdot x^{2} - 164\,in^{2} = 0

x \approx 4.528\,in

5) The equation of the box volume is:

V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x

V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x

V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}

V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}

V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}

The first derivative of the function is:

V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}

The critical points are determined by equalizing the derivative to zero:

12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0

x_{1} \approx 4.952\,in

x_{2}\approx 1.548\,in

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V'' = 24\cdot x - 78\,in

After evaluating each critical point, it follows that x_{1} is an absolute minimum and x_{2} is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

x \approx 1.548\,in

The surface area of the box is:

S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}

S = 164.830\,in^{2}

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