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Sauron [17]
2 years ago
15

8-37 better than or equal to 29

Mathematics
1 answer:
Zina [86]2 years ago
5 0

The statement that 8-37 greater than or equal to 29 is false and the correct statement is 8 -  27 is less than 29

<h3>What is an inequality?</h3>

An inequality can be defined as a relation which makes an unequal comparison between two numbers or mathematical expressions.

It is used most often to compare two numbers on the number line by their size

The information given be represented as;

8 - 37\geq 29

Now, let's find the difference from the left side

We have;

- 29\geq 29

From this, we can see that the statement is wrong as - 29 is neither greater than or equal to 29

The correct statement should be; 8 -  27 is less than 29

Thus, the statement that 8-37 greater than or equal to 29 is false and the correct statement is 8 -  27 is less than 29

Learn more about inequalities here:

brainly.com/question/24372553

#SPJ1

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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)

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