Given s (t )equals 3 t squared plus 6 ts(t)=3t2+6t, where s(t) is in feet and t is in seconds, find each of the following. a) v(t) b) a(t) c) The velocity and acceleration when tequals=22 sec

The correct question is:

Given s(t) = 3t² + 6t, where s(t) is in feet and t is in seconds, find each of the following.

a) v(t)

b) a(t)

c) The velocity and acceleration when t =22 sec

Solution:

Given s(t) = 3t² + 6t

(a) The velocity, v(t) is the derivative of s(t). That is v(t) = s'(t)

s'(t) = v(t) = 6t + 6

(b) The acceleration, a(t) is the derivative of v(t). That is a(t) = v'(t)

v'(t) = a(t) = 6

(c) When t = 22 sec,

v(t) = v(22) = 6(22) + 6 = 132 + 6

= 138

a(t) = a(22) = 6

At t = 22, the velocity is 138, and acceleration is 6.

25.2

step-by-step explanation:

2+4x6-8/10= 25.2

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.