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baherus [9]
3 years ago
11

Line segment GH is divided by point J in the ratio of 1:4. Endpoint G is at (7,5) and point J is at (10, 14).

Mathematics
1 answer:
wlad13 [49]2 years ago
7 0

Answer:

(x2 , y2) = (22,50)

Step-by-step explanation:

Given:

m1 : m2 = 1:4

(x1 , y1) = (7 , 5)

(X ,Y) = (10 ,14)

Find:

(x2 , y2)

Computation:

X = (m1x2 + m2x1) / (m1+m2)

10 = [(1)(x2) + (4)(7)]/(1+4)

10 = [x2 + 28]/5

22 = x2

Y = (m1y2 + m2y1) / (m1+m2)

14 = [(1)(y2) + (4)(5)]/(1+4)

14 = [y2 + 20]/5

70 = y2 + 20

50 = y2

(x2 , y2) = (22,50)

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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
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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

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Then we cross multiply

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If area of the red quadrilateral = 2880

Then we use the following proportion

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\frac{2880}{42}  =  \frac{x}{7}

Then we cross multiply

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