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3 years ago

What is the angle in both radians and degrees determined by an arc of length 2pi meters on a circle of radius 8 meters?

1 answer:

6 0

$\bf \textit{arc's length}\\\\
s=r\theta ~~
\begin{cases}
r=radius\\
\theta = angle~in\\
\qquad radians\\
------\\
r=18\\
s=2\pi 
\end{cases}\implies 2\pi =18\theta \implies \cfrac{2\pi }{18}=\theta\implies \cfrac{\pi }{9}=\theta \\\\
-------------------------------$

$\bf \textit{arc's length}\\\\
s=\cfrac{\theta \pi r}{180} ~~
\begin{cases}
r=radius\\
\theta = angle~in\\
\qquad degrees\\
------\\
r=18\\
s=2\pi 
\end{cases}\implies 2\pi =\cfrac{\theta \pi 18}{180}
\\\\\\
2\pi =\cfrac{\theta \pi }{10}\implies \cfrac{2\pi (10)}{\pi }=\theta \implies 20=\theta$

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3 years ago

Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera

Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = $\frac{1}{2}(\text{Base})\times (\text{Height})$

Area of ΔADC = $\frac{1}{2}(\text{CD})(\text{AD})$

= $\frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})$

= $\frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)$

= $\frac{1}{2}(\sqrt{169-144})(12)$

= $\frac{1}{2}(5)(12)$

= 30 cm²

Area of equilateral triangle II = $\frac{\sqrt{3} }{4}(\text{Side})^2$

Area of equilateral triangle II = $\frac{\sqrt{3}}{4}(13)^2$

= $\frac{(1.73)(169)}{4}$

= 73.0925

≈ 73.09 cm²

Area of rectangle III = Length × width

= CF × CD

= 7 × 5

= 35 cm²

Area of trapezium EFGH = $\frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})$

Since, GH = GJ + JK + KH

17 = $\sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}$

12 = $\sqrt{81-x^2}+\sqrt{225-x^2}$

144 = (81 - x²) + (225 - x²) + 2 $\sqrt{(81-x^2)(225-x^2)}$

144 - 306 = -2x² + $2\sqrt{(81-x^2)(225-x^2)}$

-81 = -x² + $\sqrt{(81-x^2)(225-x^2)}$

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = $\frac{1}{2}(5+17)(9)$

= 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

= 237.09 cm²

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3 years ago

An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. Step 2 of 2 : S

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Answer:

The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.