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2 years ago

9 + 4(x + 2) - 3x What is the term for best describes 3

1 answer:

4 0

An expression is defined as a set of numbers, variables, and mathematical operations. The term that best describes 3 is coefficient.

<h3>What is an Expression?</h3>

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.

The value of 3 in the expression, 9 + 4(x + 2) - 3x, is that it is the coefficient of x in the third term.

Hence, the term that best describes 3 is coefficient.

Learn more about Expression:

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6-1/2z=2/3<br> what is z?

Marina CMI [18]

6 - 1/2z = 2/3
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- 1/2z = -16/3
divide both sides by -1
1/2z = 16/3
rewrite
1z/2 = 16/3
multiply both sides by 2
1z = 32/3
z = 32/3 or 10 2/3

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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

Zarrin [17]

Answer:

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

Therefore, x = 0.53 cm and the Maximum volume = 1.75 cm³

Step-by-step explanation:

Please refer to the attached diagram:

The volume of the box is given by

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

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

The width of shaded region is given by

$Width = 3 - 2x \\\\$

The length of shaded region is given by

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

So, the volume of the box becomes,

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

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 may solve the quadratic equation using the quadratic formula.

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

The values of coefficients a, b, c are

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

Substituting the values into quadratic formula yields,

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

Volume of box when 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 box when x = 0.53:

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

As you can see, the volume of the box is maximized when x = 0.53 cm

Therefore, x = 0.53 cm and the Maximum volume = 1.75 cm³

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3 years ago