A particle moves along the x-axis so that at any time t≥0 its position is given by x(t)=12(a−t)2, where a is a positive constant. For what values of t is the particle moving to the right?

For the particle to move to the right, 0 ≤ t ≤ a

Step-by-step explanation:

The particle under motion on the x-axis given by the equation x(t) = 12(a−t)² for t ≥ 0, where a is a positive constant.

To find the values of t for which the particle is moving to the right, this implies that the particle must move along the positive x-axis. This implies that x(t) ≥ 0.

So, to find the value of t for which the particle moves to the right, we solve the inequality x(t) ≥ 0 ⇒ 12(a−t)² ≥ 0

12(a−t)² ≥ 0

(a−t)² ≥ 0

taking square root of both sides, we have

√(a−t)² ≥ √0

(a−t) ≥ 0

a ≥ t ⇒ t ≤ a

So, for the particle to move to the right, 0 ≤ t ≤ a.

i dont know it

step-by-step explanation:

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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