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3 years ago

14

Please help with this question

2 answers:

Basile [38]3 years ago

4 0

Step-by-step explanation:

m< AOC = 90°

m< AOB +m<BOC = 90°

6x-12+3x +30 = 90°

9x +18° = 90°

9x = 90- 18° = 72°

x = 8°

m<AOB =6(8)-12 = 48-12 = 36°

aksik [14]3 years ago

3 0

Since, OA is perpendicular to OC

$m\angle A OC= 90 \degree \\ m\angle AOB \: + m\angle BOC = 90 \degree \\ (6x - 12) \degree + (3x + 30) \degree = 90 \degree \\ 6x + 3x + 30 - 12 = 90 \degree \\ 9 x + 18 = 90 \\ 9x = 90 - 18 \\ 9x = 72 \\ x = \frac{72}{9} \\ x = 8 \degree$

$m\angle \: AOB = 6x - 12 \\ m\angle \: AOB = 6(8) - 12 \\ m\angle \: AOB = 48 - 12 \\ m\angle \: AOB = 36 \degree$

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Answer:

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Step-by-step explanation:

Refer the attached figure .

A surveyor moves 160 feet away from the base of the pole i.e. ED = 160 m

Referring the image .

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A transit i.e. AE is 6 feet tall.

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With a transmit the angle of elevation to the top of the pole to be 62° i.e. ∠BAC = 62°

To Find BC , use tigonometric ratio.

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan 62^{\circ}=\frac{BC}{AC}$

$Tan 62^{\circ}=\frac{BC}{160}$

$1.88*160=BC$

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

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Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

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Let $u=xy$ , the partial derivative of u wrt y is

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