A boy moves down the hall at 1.2 m/s. When he sees his arch nemesis coming, he begins to run. After 3.2 s, he is moving at 4.4 m/s. What is his acceleration?

1m/s²

Step-by-step explanation:

Acceleration is the change in velocity of a body with respect to time.

Acceleration = Change in velocity/time

Acceleration (a) = final velocity(v)-initial velocity(u)/time(t)

Given parameters:

final velocity v = 4.4m/s

initial velocity u = 1.2m/s

time taken = 3.2s

Required

Acceleration

Substitute the given parameters into the formula to get the acceleration

Acceleration = 4.4 - 1.2/3.2

Acceleration = 3.2/3.2

Acceleration = 1m/s²

Therefore the boy acceleration is 1m/s²

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)