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Xelga [282]
3 years ago
10

line segment with one endpoint at (2,5) is divided in the ratio 4:5 by the point (10,1). which coordinates could represent the o

ther endpoint.
Mathematics
1 answer:
rodikova [14]3 years ago
5 0

Answer:

(20,-4)

Step-by-step explanation:

We are given;

One end point as (2,5)

Point of division as (10,1)

The ratio of division as 4:5

We are required to calculate the other endpoint.

Assuming the other endpoint is (x,y)

Using the ratio theorem

If the unknown endpoint is the last point on the line segment;

Then;

\left[\begin{array}{ccc}10\\1\end{array}\right]=\frac{4}{5} \left[\begin{array}{ccc}x\\y\end{array}\right]+\frac{5}{9}\left[\begin{array}{ccc}2\\5\end{array}\right]

Therefore; solving the equation;

10=\frac{4}{9}x+\frac{10}{9}

solving for x

x = 20

Also

1=\frac{4}{9}y+\frac{25}{9}

solving for y

y= -4

Therefore,

the coordinates of the end point are (20,-4)

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Answer:

x = 0.53 cm

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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

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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

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

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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

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x = 0.53 cm

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