Solve 3/2^x-4=16 Two step equations

Which expression finds the measure of an angle that is coterminal with a 300° angle?

icang [17]

The expression that gives an angle that is coterminal with 300 is 300-720. Two angles are said to be coterminal if when they are drawn in a standard position, their terminal sides are on the same location. The expression gives an angle of 420 where when it is drawn the terminal sides are on the same location with the 300.

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3 0

3 years ago

Read 2 more answers

What is the equation in standard form of the line that passes through the point (-7, 2) and has a

Orlov [11]

Answer:

y=3x+23 (lmk if you need me to explain)

Step-by-step explanation:

3 0

3 years ago

Which of the following describes the behavior of the graph shown over the interval (-2,2)?

sergejj [24]

Hello!

We are trying to describe the behavior of the graph given in the question.

To help us understand how to solve this question, we would need to understand concavity.

There are two types of concavity:

Concave up
Concave down

When a graph is concave up, the slope of the line would look like a "U".

When a graph is concave down, the slope of the line would look like a "U" that is flipped upside down.

In this case, we can see that the graph is concave down.

We can tell that the slope is negative due to the fact that the slope is going down, which results in the graph having a negative slope.

We can also tell that the graph is decreasing due to the fact that the line is doing downward.

Answer:

C). negative and decreasing

7 0

1 year ago

A ball is thrown into the air with an initial velocity of 22 meters per second. The quadratic function h(x)= -4.9t square +22t+5

MrRa [10]

Answer:

Height of the ball after 3 seconds = 27.4 feet.

Step-by-step explanation:

h(t) = -4.9t^2 + 22t + 5.5

h(3) means the height of the ball at time 3 seconds after the ball was thrown.

h(3) = -4.9(3)^2 + 22(3) + 5.5

= 27.4 feet

5 0

3 years ago

The data sets show the years of the coins in two collections. Derek's collection: 1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910

KATRIN_1 [288]

Answer:

Derek's collection :

Mean= 1929

Median= 1930

Range= 54

IQR = 48

MAD= 23.75

Paul's collection:

Mean= 1929

Median= 1929.5

Range= 15

IQR = 6

MAD= 3.5

Step-by-step explanation:

1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910

Mean is given by:

(1950+1952+ 1908+1902+1955+1954+1901+1910)/8

=1929

absolute deviation from mean is:

|1950-1929|= 21

|1952-1929|= 23

|1908-1929|= 21

|1902-1929|= 27

|1955-1929|= 26

|1954-1929|= 25

|1901-1929|= 28

|1910-1929|= 19

from the mean of absolute deviation gives the MAD of the data i.e.

(21+23+21+27+26+25+28+`9)/8

23.75

:arrange the given data to get the range and median

1901 1902 1908 1910 1950 1952 1954 1955

The minimum value is: 1901

Maximum value is: 1955

Range is: Maximum value-minimum value

Range=1955-1901

Range= 54

median is (1910+1950)/2

1930

the lower set of data=

1901 1902 1908 1910

first quartile becomes

1902+1908)/2

Q1=1905

and upper set of data is:

1950 1952 1954 1955

we find the median of the upper quartile or third quartile is:

1952+1954)/2=1953

Q3-Q1=1953-1905=

IQR=48

Paul's collection:

1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920

Mean is given by:

1929+1935+ 1928+ 1930+ 1925+ 1932+1933+1920)/8

1929

absolute deviation from mean is:

|1929-1929|=0

|1935-1929|= 6

|1928-1929|= 1

|1930-1929|= 1

|1925-1929|= 4

|1932-1929|= 3

|1933-1929|= 4

|1920-1929|= 9

Hence, we get:

MAD=0+6+1+1+4+3+4+9/8

28/8

3.5

arrange the data in ascending order we get:

1920 1925 1928 1929 1930 1932 1933 1935

Minimum value= 1920

Maximum value= 1935

Range= 15 ( 1935-1920=15 )

The median is between 1929 and 1930

Hence, Median= 1929.5

Also, lower set of data is:

1920 1925 1928 1929

the first quartile or upper quartile is

1925+1928/2

1926.5

and the upper set of data is:

1930 1932 1933 1935

We have

1932+1933)/2

1932.5

IQR is calculated as:

Q3-Q1

6

7 0

3 years ago