Shota is a dangerous fellow who likes to go rock climbing in active volcanoes. one time, when he was 303030 meters below the edge of a volcano, he heard some rumbling, so he decided to climb up out of there as quickly as he could. he climbed up at a constant rate. after 4.54.54, point, 5 seconds, he was 7.57.57, point, 5 meters below the edge of the volcano. how fast did shota climb? meters per second

The answer is 5 m/s

Step-by-step explanation:

I assume that he was 30 meters below the volcano and that after 4.5 seconds, he was 7.5 meters below the edge.

That means that the distance covered by Shota was the difference between his initial position and his new position:

30 - 7.5=22.5 meters

And now, we can know how fast he was climbing by dividing the distance covered over the time he covered that distance:

So, he climbed the volcano at a constant rate of 5 m/s.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

(x+3)(x+7)

step-by-step explanation: