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2 years ago

How many radians is 144°?

Mathematics

2 answers:

aleksandrvk [35]2 years ago

4 0

Answer:

180 radians

Step-by-step explanation:

Convert from degrees to radians using the ratio

180

.4π5 radians

Vsevolod [243]2 years ago

4 0

The raidians would be 4 * Pi / 5

While your Circles are 2 / 5

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Ratio equivalent to 2:1

Nataliya [291]

Answer:

4:2

8:4

12:6

16:8

6 0

3 years ago

How do i solve v=pi/3*r (squared)h for h

Colt1911 [192]

$V=\dfrac{\pi}{3}\cdot r^2h\ \ \ \ \ |multiply\ both\ sides\ by\ 3\\\\\pi r^2h=3V\ \ \ \ |\divide\ both\ sides\ by\ \pi r^2\\\\\boxed{h=\dfrac{3V}{\pi r^2}}$

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2 years ago

The school basketball team won the game. The finals score was a two digit number. The digit in the tens place was the number tha

8_murik_8 [283]

Answer:

1? Because its the only number beetween 2 and 0.

5 0

3 years ago

after a party, you have 2/5 of the brownies you made left over. there are 16 left. how many brownies did you make for the party

vodomira [7]

16 = 2/5
16 ÷ 2 = 2/5 ÷ 2/1
8 = 1/5
8 * 5 = 1/5 * 5
40 = 5/5
You started the party with 40 brownies

7 0

3 years ago

Read 2 more answers

At a certain high school, 15 students play stringed instruments and

Aleks [24]

Answer:

0 is the probability that a randomly selected student plays both a stringed and a brass instrument.

Step-by-step explanation:

Given that:

Number of students who play stringed instruments, N(A) = 15

Number of students who play brass instruments, N(B) = 20

Number of students who play neither, N( $A \cup B$ )' = 5

To find:

The probability that a randomly selected students plays both = ?

Solution:

Total Number of students = N(A)+N(B)+N( $A \cup B$ )' =15 + 20 + 5 = 40

(As there is no student common in both the instruments we can simply add the three values to find the total number of students)

As per the venn diagram, no student plays both the instruments i.e.

$N(A\cap B) =0$

Formula for probability of an event E can be observed as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A\cap B) = \dfrac{N(A\cap B)}{N(U)}\\\Rightarrow \dfrac{0}{40}\\\Rightarrow P(A\cap B) = 0$

So, 0 is the probability that a randomly selected student plays both a stringed and a brass instrument.

5 0

2 years ago