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3 years ago

List the following in order from least steep to most steep.

Mathematics

1 answer:

klemol [59]3 years ago

5 0

Answer:

h(x), f(x), j(x), g(x)

Step-by-step explanation:

Given functions

f(x)= -3/4x
g(x)= -5x
h(x)= 5/7x
j(x)= 11/3x

To compare steepness we'll compare absolute values of slopes

The steepness increases as the slope increases

The slopes are

3/4, 5, 5/7, 11/3

Comparing the fractions

3/4 = 21/28, 5/7 = 20/28, so 3/4 > 5/7

And

11/3 < 4

So the order is

5/7 < 3/4 < 11/3 < 5

As functions, the order is

h(x), f(x), j(x), g(x)

You might be interested in

Evaluate the expression for b = 0 and c= -11. bc + b

avanturin [10]

Answer:

If b = 0 and c = -11, all you have to do is plug in the values into bc + b:

Step-by-step explanation:

(0)(-11) + 0 = 0

3 0

2 years ago

Read 2 more answers

Graph a line with a slope of 4 that contains the point (3,0).

Korvikt [17]

Answer:

just graph the line: y=4x-12

Step-by-step explanation:

4 0

3 years ago

Workers have been down the street installing two blinking red lights in front of the fire station. The two lights are always tur

daser333 [38]

Answer:

After 88 seconds, they will blink together.

Step-by-step explanation:

Time after which the first light blinks = 8 seconds

Time after which the second light blinks = 22 seconds

Initially, both the lights blink together.

To find:

The time at which they will blink together again.

Solution:

Let us have a look at the time after which each light will blink.

First light will blink after:

8 s, 16 s, 24 s, 32 s, 40 s, 48 s, 56 s, 64 s, 72 s, 80 s, 88 s, 96 s, 104 s, ......

Second light will blink after:

22 s, 44 s, 66 s, 88 s, 110 s, .....

It can be seen that common time is 88 second, at which the both will blink together.

So, the answer is 88 s.

This answer can also be computed by taking the LCM (Least Common Multiple) of both the numbers i.e. 8 and 22.

Using the factorization method:

$8=\underline{2}\times 2\times 2$

$22=\underline{2}\times 11$

Common is 2.

Therefore the LCM will be:

$2\times 2\times2\times 11 = \bold{88}$

So, the answer is 88 s.

5 0

3 years ago

1948 Olympics

mash [69]

Answer:

For 1948 Men's :Interquartile range is 1.5, Median is 58.3

For 2012 Men's: Interquartile range is 0.315, Median is 47.86

You can infer that in 2012 the swimmers were better and it was more competitive as the interquartile range was lower and the median was also lower as well

Step-by-step explanation:

1948 Men's 100m

57.3, 57.8, 58.1, 58.3, 58.3, 59.3, 59.6,1:00.5

⬆ ⬆ ⬆

Low IQR Median High IQR

Low IQR is average of 57.8 and 58.1 = 57.95

High IQR is average of 59.3 and 59.6 = 59.45

Median is average of 58.3 and 58.3 = 58.3

Interquartile range is High IQR- Low IQR

59.45 - 57.95=1.5

2012 Men's 100m

47.52, 47.53, 47.8, 47.84, 47.88, 47.92, 48.04, 48.44

⬆ ⬆ ⬆

Low IQR Median High IQR

Low IQR is average of 47.53 and 47.8 = 47.665

High IQR is average of 48.04 and 47.92 = 47.98

Median is average of 47.88 and 47.84 = 47.86

Interquartile range is High IQR-Low IQR

47.98-47.665=0.315

Read more on Brainly.com - brainly.com/question/14915771#readmore

8 0

3 years ago

ACDF,BE is a mid segment what is x?

Ksenya-84 [330]

Answer:

X= 15

Step-by-step explanation:

the above equation will be used to determine the value of x.

6x-12= 2x+20+18

6x-2x = 20+12+18

4x= 60.

X= 60/4

X= 15

x = 15

7 0

3 years ago