Witch equation is it??

I am Lyosha [343]

Central angle = 360 / 24 = 15 degrees

ARc length = r * theta, where theta is the angle in radians and r = radius
= 15 * 15 * pi/180
= 3.93 feet to nearest hundredth.

Area of sector = area of circle / 24 = pi*15^2 / 24 = 29.45 ft^2

6 0

3 years ago

Q7Which statement is incorrect for the graph of thefunction y = -5 cosGcos( 3 (x – 3)) +62The period is 4.The amplitude is 3.The

guajiro [1.7K]

The given function is,

$y=-5\cos (\frac{\pi}{2}(x-3))+6$

The graph can be drawn as,

The period will be distance between two consequetive maximas,

$\text{Period}=1-(-3)=4$

Thus, period is correct.

The amplitude can be determined as,

$\begin{gathered} Amplitude=\frac{\max -\min }{2} \\ =\frac{11-1}{2} \\ =5 \end{gathered}$

Thus, amplitue is incorrect.

The range is,

$y\in\lbrack1,11\rbrack$

Thus, the range is correct.

The midline is,

$\begin{gathered} \text{midline}=\frac{\max +\min }{2} \\ =\frac{1+11}{2} \\ =6 \end{gathered}$

Thus, the midline is correct.

Thus, option (b) is the solution.

5 0

1 year ago

There is 60% chance that it will rain today. If it rains, the probability of watching a movie is 0.5. But, if it does not rain t

agasfer [191]

Applying the given probabilities, it is found that there is a 0.42 = 42% probability of watching a movie today.

These following <em>probabilities </em>involve a movie being watched today:

0.5 of 0.6(if it rains today).

0.3 of 1 - 0.6 = 0.4(if it does not rain today).

Hence, adding them:

$p = 0.5(0.6) + 0.3(0.4) = 0.3 + 0.12 = 0.42$

0.42 = 42% probability of watching a movie today.

A similar problem is given at brainly.com/question/14398287

5 0

3 years ago

there is 3/5 of pie left if 4 friends wanted to share the pie equally how much pie would each friend receive/7095668/7abc4acb?ut

Serjik [45]

You would divide 3/5 by 4/1 so then that would mean you keep change flip so its 3/5 times 1/4 which is equal to 3/20 each person or friend would get

3 0

3 years ago

Which of the following are true statements.

Mariulka [41]

Answer:

Second statement is true.

The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

Step-by-step explanation:

for first part of statement

The lengths 7, 40 and 41 can not be sides of a right triangle.

If the square of long side is equal to the sum of square of other two sides

then the given length can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

$c^{2} =a^{2} +b^{2}$ ----------(1)

Let $c=41$ and $a = 7$ and $b=40$

Put all the value in equation 1.

$41^{2} =7^{2} +40^{2}$

$1681=49+1600$

$1681=1649$

Therefore, the square of long side is not equal to the sum of square of other two sides, So given lengths 7, 40 and 41 can not be sides of a right triangle.

for second part of statement.

The lengths 12, 16, and 20 can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

Let $c=20$ and $a = 12$ and $b=16$

$20^{2} =12^{2} +16^{2}$

$400=144+256$

$400=400$

Therefore, the square of long side is equal to the sum of square of other two sides, So given the lengths 12, 16, and 20 can be sides of a right triangle.

Therefore, The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

8 0

3 years ago