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2 years ago

What is the measure (in radians) of central angle in the circle below?

Mathematics

1 answer:

kati45 [8]2 years ago

3 0

Answer:

∅ = 5 radians

Step-by-step explanation:

$Arc.Length=2\pi r(\frac{Ctrl.Angle}{360} )$

$Ctrl.Angle=\frac{360(Arc.length)}{2\pi r}$

∅ $=\frac{(360)(15)}{2\pi (3)}$

∅ = 286.48°

In radians:

∅ = (286.48)(3.1416)/180 = 5 radians

Hope this helps

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What is the value of 3 + (-1/4)² ÷ 1/2? A. 6 1/8 B. 21 1/8 C. 7 D. 3 1/8

Amiraneli [1.4K]

3 + (-1/4)² ÷ 1/2
= 3 + 1/16 * 2
= 3 + 1/8
= 3 1/8

answer

D. 3 1/8

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3 years ago

there are 30 marked seals. In a survey, you count 58 seals, 16 of which were marked. About how many seals are in the area? Round

Rina8888 [55]

Answer:

109 seals

Step-by-step explanation:

Let

x -----> approximate total seals in the area

we know that

In a survey, you count 58 seals, 16 of which were marked

Using proportion

Find out how many seals are in the area , if there are 30 marked seals

$\frac{58}{16} =\frac{x}{30}\\\\x=58(30)/16\\\\x=108.75\ seals$

Round to the nearest whole number

$108.75\ seals=109\ seals$

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3 years ago

What are the solutions to the quadratic equation (5y + 6)2 = 24?

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4 years ago

Solve ln 2 + ln x = 5 . Round to the nearest thousandth, if necessary. show work please

Paraphin [41]

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3 years ago

The length of a rectangle is 5 inches more than its vedth. The area of the rectangle is equal to

jarptica [38.1K]

Answer:

The width = x = 14

The length = x+5 = 14+5 = 19

Step-by-step explanation:

Let l = length, w = width, A = area = l×w , and P = perimeter = 2(l+w)

Let 'x' be the width

As the length 'l' of a rectangle is 5 inches more than its width.

so the length will be = x+5

As the Area of the rectangle is equal to 2 inches more than 4 times the perimeter.

A = 4P + 2

so the equation becomes

l × w = 4×2(l+w)+2

substituting w=x, l = x+5,

(x+5)x = 8(x+5+x)+2

x²+5x = 8(5+2x)+2

x²+5x = 40+16x+2

x²+5x = 16x+42

x²+5x-16x-42 =0

x²-11x-42=0

$x^2-11x=42$

$\mathrm{Add\:}a^2=\left(-\frac{11}{2}\right)^2\mathrm{\:to\:both\:sides}$

$x^2-11x+\left(-\frac{11}{2}\right)^2=42+\left(-\frac{11}{2}\right)^2$

$x^2-11x+\left(-\frac{11}{2}\right)^2=\frac{289}{4}$

$\left(x-\frac{11}{2}\right)^2=\frac{289}{4}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{11}{2}=\sqrt{\frac{289}{4}}$

$x-\frac{11}{2}=\frac{\sqrt{289}}{\sqrt{4}}$

$x-\frac{11}{2}=\frac{\sqrt{289}}{2}$

$x-\frac{11}{2}=\frac{17}{2}$

$x-\frac{11}{2}+\frac{11}{2}=\frac{17}{2}+\frac{11}{2}$

$x=14$

similarly solving

$x-\frac{11}{2}=-\sqrt{\frac{289}{4}}$

$x-\frac{11}{2}=-\frac{17}{2}$

$x-\frac{11}{2}+\frac{11}{2}=-\frac{17}{2}+\frac{11}{2}$

$x=-3$

x = 14, or x = -3

As the width 'x' can not be negative.

so x = 14

Thus,

The width = x = 14

The length = x+5 = 14+5 = 19

6 0

3 years ago