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2 years ago

9

I had to put two screenshot because it would screenshot the whole question but plz help

1 answer:

Tcecarenko [31]2 years ago

3 0

Answer:

i think its D dont fullly take my word tho

Step-by-step explanation:

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For the given central angle, determine the distance traveled along the unit circle from the point (1, 0). 360 degrees a. 6.28 un

Alekssandra [29.7K]

Answer:

The length of a 180° arc of a unit circle is π ≈ 3.14 units.

Step-by-step explanation:

Use your knowledge of the circumference of a circle (the length full around) and the fact that there are 360° in the central angle of a full circle. The distance around is proportional to the angle, so an arc of measure 180° will have a length equal to

... (180°/360°) × circumference = (1/2)×circumference

For a unit circle, the circumference is 2π (= π×diameter = 2π×radius). Half that length is π units.

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2 years ago

SOMEONE PLEASE HELP!!!!!!!!<br> (Also please don't reply if you don't know)

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Step-by-step explanation:

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2 years ago

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

8 0

2 years ago

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Question 19: Please help. What is the equation of the line, in slope-intercept form, that is perpendicular to the line with the

alukav5142 [94]

Answer:

the first optionnnnnnn

7 1

2 years ago

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(20 PTS) Please help!!!! I've been stuck for 1 hour already! :(

Sindrei [870]

We have: 10 + 7 = 17

Answer: 17

Ok done. Thank to me :>

7 0

1 year ago