Please help me with this question and please show work I appreciate it

What is the area of the sector ?

worty [1.4K]

They forget to say "not to scale". I'm guessing this is trig because I don't see another way to do it.

Let's consider x=chord PQ first. By the Law of Cosines

$x^2 = 3^2 + 5^2 - 2(3)(5) \cos 81^\circ = 34 - 30\cos 81^\circ$

We have an isosceles triangle formed by two radii of 9 cm and x=PQ. By the Law of Cosines again,

$x^2 = 9^2 + 9^2 - 2(9)(9) \cos B$

$x^2 = 162 - 162 \cos B$

$\cos B = \dfrac{162 - x^2}{162} = 1 - \dfrac{x^2}{162}$

$\cos B = 1 - \dfrac{34 - 30\cos 81^\circ}{162}$

$B = \arccos \left(1 - \dfrac{34 - 30\cos 81^\circ }{162} \right)$

The area of the circle is the fraction given by the angle,

$A = \dfrac{B}{360^\circ} \pi r ^2$

$A \approx \dfrac{ \arccos \left(1 - \dfrac{34 - 30\cos 81^\circ}{162} \right) }{360} (3.142)9^2$

$A\approx 24.7474332960707$

Answer: 24.7 sq cm

8 0

3 years ago

A student scored 66,87,and 97 on three algebra tests. What must he score on the fourth test in order to have an average grade of

adoni [48]

Answer:66,97

Step-by-step explanation:

8 0

3 years ago

For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of water around the boat, ????(????),

Effectus [21]

Answer:

a. Period, T = 12.57 hours. b. Latest time = 294.16 hours, 6 a.m 12 days later

Step-by-step explanation:

For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of water around the boat, d(t), in meters, where t is measured in hours since midnight, is d(t) = 5 + 4.6sin(0.5t). (a) What is the period of the tides in hours? (b) If the boat leaves the bay at midday, what is the latest time it can return before the water becomes too shallow?

a. The period, T of the tides is gotten from ω = 2π/T, where ω is the angular frequency of the wave and T its period. Comparing the sine part of equation of the tides with the general equqtion of a sine wave, Asinωt, thus Asinωt = 4.6sin(0.5t),

so ω = 0.5 rad/s.

Equating this to 2π/T implies

0.5 = 2π/T

therefore, T = 2π/0.5 = 12.566 hours ≈ 12.57 hours

b. The tide becomes too low if it is less than 2.5 m i.e d(t) < 2.5 m. So, equating d(t) as 2.5 in the tidal equation, we have d(t) = 5 + 4.6sin(0.5t).

2.5< 5+ 4.6sin(0.5t)

2.5-5 < 4.6sin(0.5t)

-2.5 < 4.6sin(0.5t)

-2.5/4.6 < sin(0.5t)

-0.543 < sin(0.5t)

sin⁻¹(-0.543)<0.5t

-32.92<0.5t. Since we cannot have negative time, we add 180 to -32.92 to give 147.08

147.08<0.5t

t>147.08/0.5=294.16 hours

So, if t is greater than 294.16 hours, the water will be shallow. That is, below 2.5 m. which is 12 days 6hours 8 min 38 s which is 6 a.m 12 days later

3 0

3 years ago

What is the answer to this?

lesantik [10]

Answer:

c. positive w/ 1 outlier

Step-by-step explanation:

see picture

4 0

3 years ago

Consider the following graph, which details the observed bear population in two national parks. A graph titled Bears Observed in

damaskus [11]

Answer:

b. II and III

Step-by-step explanation:

A graph is a data presentation that has two axis - x and y, divided into appropriate scales.

A scale is used to divide given units into equal parts. It is crucial to and used to create uniformity in drawing of graphs. Choosing a good scale to represent an information on a graph is vital to avoid confusion. This can be done by considering the variations in the values forming the data.

In the question, the single graph shows an information about the bear population in two national parks. Thus, the scale of the y-axis must be adjusted and the identity of the two parks be more differentiated so as to make the graph less misleading.

6 0

3 years ago