I GIVE BRAINLIKLSTTT

Rina8888 [55]

Answer: 2.5, 5.4

Step-by-step explanation:

$-16t^2 +126t=213\\\\16t^2 -126t+213=0\\\\t =\frac{-(-126) \pm \sqrt{(-126)^2 -4(16)(213)}}{2(16)}\\\\t \approx 2.5, 5.4$

8 0

1 year ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

Jonathan’s class has 30 boys. Of the students in his class, 60% are girls. How many girls are in Jonathan’s class?

raketka [301]

Answer:

45 girls

Step-by-step explanation:

Remark
There are only 2 choices for gender. So if the class is 60% girls, the there must be 100 - 60 = 40% for the boys.
But we are told that the boys are 30 in number.
Let x = the total number of students.

Solution

40/100 * x = 30 Change the % to a decimal
0.4 x = 30 Divide both sides by 0.4
0.4 x / 0.4 = 30 / 0.4 Do the division
x = 30/0.4
x = 75 students in total

Formula

Total of Boys and Girls = Number of boys + number of girls.

Givens

Total = 75
Number of boys = 30

Solution II

75 = x + 30 Subtract 30 from each side
75 - 30 = x +30 - 30
45 = x
There are 45 girls in the class.

5 0

3 years ago

the diameter of your bycicle wheel is 25 inches how far will you move in 2 turns of your wheel use 3.14 for pi

earnstyle [38]

Answer:

157 in

Step-by-step explanation:

2x3.14x25 = 157 in

3 0

2 years ago

Kaylee's history class quiz scores for the current grading period are 10, 50, 75, 80, 82,

goldfiish [28.3K]

B is the answer I need the points

5 0

3 years ago