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denis23 [38]
3 years ago
11

the measure of angle e, the angle of elevation from point a to point b, is (3x+1). The measure of angle s, the angle of depressi

on from point b to point a, is 2(x+8). Find the measure of each angle.

Mathematics
2 answers:
bagirrra123 [75]3 years ago
7 0
C    m angle E= 46 degrees,  m angle D=46 degrees
Shalnov [3]3 years ago
3 0

Answer:   m∠ E = 46°, m∠ S = 46°

Step-by-step explanation:

Angle of elevation from point a to b ( angle E ) is the angle between the horizontal and the line from a to b .

And, Angle of depression from b to a ( angle S ) is the angle from the horizontal downward from b to a,

Since, all the horizontal lines are parallel to each other,

By the alternative interior angle theorem,

Angle of elevation from point a to b = Angle of depression from point b to a,

⇒ m ∠ E = m ∠ S  

⇒ 3x + 1 = 2(x+8)  ( Given )

⇒ 3x + 1 = 2x + 16

⇒ x = 15,

Hence, m∠ E = 3x + 1 = 45 + 1 = 46°

⇒ m∠S = 46°

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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.

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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 \\\\

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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

Maximum volume = 1.75 cm³

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Answer:

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