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vovangra [49]
3 years ago
6

Simplify -3 square root of 84x^3

Mathematics
1 answer:
morpeh [17]3 years ago
4 0
-6x (square root)21x

Hope this helps :)
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Its  {3, 3, 6, 7, 9}

because the difference in values of 6  and the other numbers are the same as difference between 5 and the other numbers in { 2, 2, 5, 6, 8}

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Slove the following <br>equation without with out transporting I)5y +10=4y-10​
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The answer is y= -20
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Help me please!!<br> hurry
zubka84 [21]

Answer:

n-3

Step-by-step explanation:

for every input, the output is three less (4-3=1, 7-3=4, 8-3=5). That rule needs to stay consistent, so no matter what the input is (n), the output is always going to be three less than, making it y or output=n-3

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3 years ago
A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six
kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

V = Length \times Width \times Height \\\\

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

Width = 3 - 2x \\\\

The length of the shaded region is given by

Length = \frac{1}{2} (5 - 3x) \\\\

So, the volume of the box becomes,

V =  \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V =  \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V =  \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V =  \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

a = 18 \\\\b = -38 \\\\c = 15 \\\\

x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 +  19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\

Volume of the box at x= 1.59:

V =  \frac{1}{2} (5 – 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\

Volume of the box at x= 0.53:

V =  \frac{1}{2} (5 – 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3

The volume of the box is maximized when x = 0.53 cm

Therefore,

x = 0.53 cm

Maximum volume = 1.75 cm³

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Measurements of 30 I believe yw
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