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3 years ago

An arc on a circle measures 295°. The measure of the central angle, in radians, is within which range?

Mathematics

2 answers:

andrezito [222]3 years ago

3 0

Answer:

3pi/2 to 2π radians

Step-by-step explanation:

The centeal angle also has 295⁰. This value is between 275⁰ (or 3×pi/2) and 360⁰ (or 2×pi).

So, the answer is: 3pi/2 to 2π radians

rodikova [14]3 years ago

3 0

Answer:

The 4 option

Step-by-step explanation:

3pi/2 to 2π radians

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Evaluate each trigonometric equation for θ.

liberstina [14]

Answer:

sinФ = 8/17

cosФ = 15/17

tan Ф = 8/15

csc Ф = 17/8

sec Ф = 17/15

cot Ф = 15/8

Step-by-step explanation:

Let us revise the trigonometry functions

Sin(x) = opposite/hypotenuse

Cos(x) = adjacent/hypoteouse

Tan(x) = opposite/adjacent

Csc(x) = hypotenuse/opposit

Sec(x) = hypotenues/adjacent

Cot(x) = adjacent/opposite

In the given figure

The opposite side to angle Ф = 8

The adjacent side to angle Ф = 15

Find the hypotenuse using Pythagoras' theorem

$Hypotenuse = \sqrt{8^{2}+15^{2} }$

$Hypotenuse = \sqrt{64+225}$

$Hypotenuse = \sqrt{289}$

$Hypotenuse = 17$

Let us use the rules above to find the trigonometry functions

sinФ = 8/17

cosФ = 15/17

tan Ф = 8/15

csc Ф = 17/8

sec Ф = 17/15

cot Ф = 15/8

4 0

3 years ago

1.) Consider the right triangle below in which a = 5 and b = 5.

patriot [66]

Answer:

c=7.07

Step-by-step explanation:

c²=a²+b²

c²=5²+5²

c²=25+25

c²=50

c=√50

c=7.07

B. c=7.07 (to nearest hundredth)

C. c=√50= 5√2 in radical form

3 0

2 years ago

Solving these quadratics for x using complete the square 3x^2-4x-1=0. Show your work.

UkoKoshka [18]

Answer: x = (sqrt(7) + 2)/3 and
x = ( – sqrt(7) + 2)/3

Explanation:

3x^2 - 4x - 1 = 0

Divide both sides by 3:

3x^2/3 - 4/3x - 1/3 = 0/3
x^2 - 4/3x - 1/3 = 0
x^2 - 4/3x = 1/3
x^2 - 4/3x + (2/3)^2 = 1/3 + (2/3)^2
(x - 2/3)^2 = 1/3 + 4/9
(x - 2/3)^2 = 7/9

Sqrt both sides:

x - 2/3 = sqrt (7/9)
x - 2/3 = |sqrt(7)/3|

Set x -2/3 = sqrt(7)/3

=> x = (sqrt(7) + 2)/3

Set x - 2/3 = - sqrt(7)/3

=> x = ( - sqrt(7) + 2)/3

6 0

3 years ago

2g-h/f=5 solve for g

LiRa [457]

You can put this solution on YOUR website!
2g-m=5-gh solve for g
2g+gh=5+m
g(2+h)=5+m
g=(5+m)/(2+h)
Cheers,
Stan H.

4 0

3 years ago

The third, fifth and eighth terms of an AP are the first 3 consecutive terms of a GP. Given that the first term of the AP is 8,

marshall27 [118]

Answer:

Common Difference = 2.

Step-by-step explanation:

An AP can be written as a1, a1 + d, a1 + 2d, a1 + 3d, a1 + 4d, a1 + 5d, a1 + 6d , a1 + 7d.

where a1 = first term and d is the common difference.

Here first term = a1 = 8

3rd term = a1 + 2d = 8 + 2d

5th term = a1 + 4d = 8 + 4d

8th term = 8 + 7d

First 3 terms of a GP are a , ar and ar^2

So from the given information:

a = 8 + 2d

ar = 8 + 4d

ar^2= 8 + 7d

Dividing the second equation by the first we have

r = (8 + 4d)/(8 + 2d)

Dividing the third by the second:

r = (8 + 7d) / (8 + 4d)

Therefore, eliminating r we have:

(8 + 4d)/(8 + 2d) = (8 + 7d)/(8 + 4d)

(8 + 4d)^2 = (8 + 2d)(8 + 7d)

64 + 64d + 16d^2 = 64 + 72d^ + 14d^2

2d^2 - 8d = 0

2d(d^2 - 4) = 0

2d = 0 or d^2 = 4, so

d = 0, 2.

The common difference can't be zero so it must be 2.

8 0

3 years ago