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Mumz [18]
2 years ago
13

Yayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

Mathematics
1 answer:
Umnica [9.8K]2 years ago
7 0

Answer:

It b

Step-by-step explanation:

We know this because 1/2 is 0.5 while 2/3 is 0.67. With this we can say that 1/2 is least than 2/3

Hope this Helped

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First answer gets brain
Alex787 [66]
3 \frac{4}{7}m =8 \frac{1}{3} \\ \\\frac{25}{7}m=\frac{25}{3} \\  \\ m= \frac{25}{3}: \frac{25}{7}= \frac{25}{3}* \frac{7}{25}= \frac{7}{3}=2 \frac{1}{3}
4 0
3 years ago
Read 2 more answers
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
Evaluate. 5x + 2y2-y3; x =2 and y=4
Elanso [62]
5x+2y2-y3:x=2 and y=4

5(2)+2(4)2-(4)3

10+8(2) -12

10+16-12

26-12

=14
7 0
3 years ago
How do you use a number line with ratios?​
Natalija [7]

Answer:

act as if the ratio is a fraction 4:3 would be 4/3..

Step-by-step explanation:

5 0
2 years ago
CD is a perpendicular bisector of chord AB and a chord through CD passes through the center of a circle. Find the diameter of th
Keith_Richards [23]

Answer:

12\frac{1}{5} in

Step-by-step explanation:

I don't know if you got this pic but this is the answer for this pic

5 0
2 years ago
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