All categories

3 years ago

Find one length of the square (7x-3) 3(x+1)

Mathematics

1 answer:

Svetllana [295]3 years ago

6 0

Area of a square = Side × Side

Side = 7x-3

Area of the square = (7x-3) × (7x-3)

= 7x(7x-3) -3(7x-3)

= 49x^2 -21x -21x +9

=49x^2 -42x +9

You might be interested in

Use the graph of the following function (x) to find the value. <br> f(1)

kvv77 [185]

From the graph of the given function , the value of f(1) = -1.

As given in the question,

From the graph of the given function,

Two coordinates from the graph are as follow:

( x₁ , y₁) = (1, -1)

( x₂ , y₂ ) = ( 0, -3 )

Equation of the line representing the function is given by:

(y - y₁) /(x-x₁) = ( y₂ -y₁)/ (x₂ -x₁)

⇒(y +1)/ (x-1) = (-3 +1)/ (0-1)

⇒ (y +1)/ (x-1) = 2

⇒y +1 = 2x -2

⇒ y = 2x -3

To get the value of x we have,

y = f(x)

⇒f(x) = 2x -3

⇒f(1) = 2(1) -3

⇒f(1) = -1

Therefore, from the graph of the given function , the value of f(1) = -1.

Learn more about graph here

brainly.com/question/17267403

#SPJ1

3 0

1 year 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

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

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