Ask question

Ask question

All categories

3 years ago

8

a red string of holiday lights blinks once every 3 seconds while a string of blue lgihts blink once every 4 seconds. how many ti

mes with both sets of lights blink at the same time in 1 minute

1 answer:

gavmur [86]3 years ago

6 0

The highest common factor of 4 and 3 is 12
one minute has 60 secs
thus it will blink 60/12=5 times

You might be interested in

The slope of (1, 5), (–3, 0) and (–2, 4), (3, 0)

chubhunter [2.5K]

Answer:

The slope of (1,5) and (-3,0) is 5/4

the slope of (-2,4) and (3,0) is -4/5

Step-by-step explanation:

5 0

3 years ago

an actor gains 20pounds for a part and then losses 15 pounds during the filming of the movie to go along with the story.who do y

guajiro [1.7K]

Let the actor's initial weight be 'x'

He is then gaining 20 pounds. The clue is in the word 'gaining' as it means 'increasing' or in Mathematics can be related as 'plus'

The expression is: x+20

He then loses 15 pounds. The clue is in the word 'losses' as it means 15 pounds is to be subtracted from the last value

Last value is x + 20
The weight now is x + 20 - 15 = x + 5

7 0

3 years ago

PLEASE HELP FAST!

svetlana [45]

Answer:

B

Step-by-step explanation:

y × y = $y^{2}$

y × -3 = -3y

$y^{2} -3y$

4 0

3 years ago

Read 2 more answers

How do you solve X/14=-21

Ket [755]

You multiply -21 by 14 to find x, and that would be -21 x 14 = -294, and that is your answer.

8 0

3 years ago

Joshua started cycling at 5:15 pm By 8:09 pm, he has covered a distance of 7250 mWhat was Joshua's average speed during that tim

vivado [14]

The average speed of Joshua during that time is 2500 m/h.

Explanation:

It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.

The total time taken by Joshua from 5:15 pm to 8:09 pm is

$2 { h } 54 \text { min }=2 \mathrm{h}+\frac{54}{60} {h}$

Dividing we get,

$2 { h } 54 \text { min }=2 \mathrm{h}+0.9 {h}$

Adding, we have,

$2 { h } 54 \text { min }=2.9 {h}$

Thus, the total time taken by Joshua is $2.9 {h}$

To determine the average speed we use the formula,

$speed=\frac{distance}{time}$

where $distance=7250$ and $time=2.9$

Hence, substituting the values we have,

$speed=\frac{7250}{2.9}$

Dividing, we get,

$speed=2500$

Thus, the average speed of Joshua during that time is 2500 m/h.

5 0

3 years ago