All categories

2 years ago

Factor 16p4 - 24p3. find the factor

Mathematics

1 answer:

Stolb23 [73]2 years ago

5 0

Answer:

8p^3 ( 2p -3)

Step-by-step explanation:

16p^4 - 24p^3

8*2* p^3 *p - 3*8*p^3

Factor out the common terms

8p^3 ( 2p -3)

You might be interested in

20. What is the perimeter of a square that has a side length of 4x - 5y? (Show Work)

sammy [17]

Answer= 16x -20y

Perimeter of a square= 4( Side length)
Therefore, 4(4x -5y)
= 16x -20y
I think that is it cuz e can’t solve it any further

8 0

3 years 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

Solve this: (a^3-2a^2)-(3a^2-4a^3)

Jobisdone [24]

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

6 0

3 years ago

Find the LCm for each set of numbers 32 128

allsm [11]

The LCM is of 128 and 32 is 128

7 0

2 years ago

Read 2 more answers

In triangle STU, s = 9 cm, t = 15 cm, and m angle U = 37 degrees. Find u.

liq [111]

We can solve the length of u using Laws of Cosine since we are given with lengths of two side and one angle.
The solution is shown below:
cos U =( s²+t²-u²)/2st
cos 37°= (9²+15²-u²)/2*9*15
0.79863*270=306-u²
u²=306-215.63
u=9.5 units
Therefore, the length of u is 9.5cm.

4 0

3 years ago

Read 2 more answers