<img src="https://tex.z-dn.net/?f=5%20%7Bx%20%3D%20125%7D%5E%7B%3F%7D%20" id="TexFormula1" title="5 {x = 125}^{?} " alt="5 {x =

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Varvara68 [4.7K]

3 years ago

125}^{?} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Novay_Z [31]3 years ago

4 0

Answer:

x = 5

Step-by-step explanation:

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Graphing Linear Inequality<br><br> y≤ 3/5x-5

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3 years ago

Read 2 more answers

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Lilit [14]

Answer:

1) 2m + 3m2 - 4m=7

7) 3m2 – 2m + 4m= 11

8) 20 + 109 + 39 - 4= 164

3) 2m + 4m - 3m2= -3

9) 4xy + x + 2xy= 0

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Step-by-step explanation:

5 0

3 years ago

Pls Help I’m stuckThere are 89 students in the Oakwood Middle School band. The number of boys in the band is 5 more than twice t

Rasek [7]

Answer:

Step-by-step explanation:

5 more than 2 times the girls

89=5+2x

x=42

4 0

2 years ago

PLZZ! HELP WITH THIS! WILL GIVE BRAINLLEIST! PROMISE! AND 11 POINTS! FOR A SIMPLE ANSWER! THANKS TO ANYONE WHO ANSWERS THIS! HUR

Alenkinab [10]

Answer:

65.52 square feet

Step-by-step explanation:

to find area of a rhombus with the diagonals multiply them together and divide by two.

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7 0

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

7 0

3 years ago