What is the distance between the following points?

Can someone help me with this and show the step to solving it

anyanavicka [17]

Answer:x = 9/16

Step-by-step explanation:

2x + 3 = 6x - 4

/3

First you gotta get rid of the divided by 3

2x / 3 = 2/3x and 3 / 3 = 1

Now you have

2/3x + 1 = 6x - 4

Second you have to get rid of the 2/3x on the left side

2/3x + 1 = 6x - 4

-2/3x -2/3x

1 = 5 1/3x - 4

Third you have to get rid of the -4

1 = 5 1/3x - 4

-4 -4

-3 = 5 1/3x

Fourth you have to divide by 5 1/3

-3 = 5 1/3x

/ 5 1/3 / 5 1/3

x = 9/16

3 0

3 years ago

Which answers describe the shape below? Check all that apply.

serg [7]

Answer:

Trapezoid and Quadrilateral

Step-by-step explanation:

Trapezoid because it has four sides that are not equal You may not be familiar with this figure because most trapezoids don't look like this. It is a quadrilateral because it is a figure that has 4 sides.

7 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

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

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