Ask question

Ask question

All categories

3 years ago

11

A football is kicked with a speed of 18.0 m/s at an angle of 36.9° to the horizontal.

1 answer:

nata0808 [166]3 years ago

8 0

Answer:

A

Step-by-step explanation:

horizontal component=18cos 36.9°≈14.39 m/s≈14.4 m/s

Vertical component=18 sin 36.9°≈10.81 m/s≈10.8 m/s

You might be interested in

-4/3,4/9,-4/27,4/81,-4/243...<br><br> what is the common ratio! HELP ASAP

kaheart [24]

Answer: The common ratio is -1/3

-----------------------------------------------------------

Work Shown:

Pick any term to divide it by the previous term. I'm going to pick the second term and divide it by the first term.

r = common ratio
r = (term 2) divided by (term 1)
r = (4/9) divided by (-4/3)
r = (4/9) * (-3/4)
r = (4*(-3))/(9*4)
r = -12/36
r = -1/3
The common ratio is -1/3

This means you multiply each term by -1/3 to get the next term

For instance
term 2 = r*(term 1)
term 2 = (-1/3)*(-4/3)
term 2 = ((-1)*(-4))/(3*3)
term 2 = 4/9

3 0

3 years ago

The ratio of players to coaches at football camp was 15 to 2. If there were 120 players, how many

Marta_Voda [28]

Answer:

16

Step-by-step explanation:

120 divided by 15 is 8

8 x 2 = 16

There are 16 coaches.

8 0

3 years ago

Coach Evans recorded the height, in inches of each player on his team. The results are shown.

marysya [2.9K]

Answer:

3

Step-by-step explanation:

Given:

Team heights (inches):

61, 57, 63, 62, 60, 64, 60, 62, 63

To find: IQRs (interquartile ranges) of the heights for the team

Solution:

A quartile divides the number of terms in the data into four more or less equal parts that is quarters.

For a set of data, a number for which 25% of the data is less than that number is known as the first quartile $(Q_1)$

For a set of data, a number for which 75% of the data is less than that number is known as the third quartile $(Q_3)$

Terms in arranged in ascending order:

$57,60,60,61,62,62,63,63,64$

Number of terms = 9

As number of terms is odd, exclude the middle term that is 62.

$Q_1$ is median of terms $57,60,60,61$

Number of terms (n) = 4

Median = $\frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th} }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{60+60}{2}=\frac{120}{2}=60$

So, $Q_1=60$

So, 25% of the heights of a team is less than 60 inches

$Q_3$ is the median of terms $62,63,63,64$

Median = $\frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th} }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{63+63}{2}=\frac{126}{2}=63$

So, $Q_3=63$

So, 75% of the heights of a team is less than 63 inches

Interquartile range = $Q_3-Q_1=63-60=3$

The interquartile range is a measure of variability on dividing a data set into quartiles.

The interquartile range is the range of the middle 50% of the terms in the data.

So, 3 is the range of the middle 50% of the heights of the students.

4 0

3 years ago

Please help im on dead line

Serjik [45]

Answer: C

Step-by-step explanation:

It is C because you are adding money to something you already have, hence positive integer

8 0

3 years ago

Factor 6(2p q)^2 -5(2p q) - 25

kherson [118]

Treat the 2p+q as x

6x2-5x-25
trial and error
(2x-5)(3x+5)
replace x with (2p+q)
(2(2p+q)-5)(3(2p+q)+5)
(4p+2q-5)(6p+3q+5)

3 0

3 years ago