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3 years ago

Please help ASAP. Find the volume of the trapezoid as a simplified expression. If correct I will give brainliest.

Mathematics

1 answer:

Leokris [45]3 years ago

6 0

Answer:

2x^2*(2x^3+8x^2+2x) or

4x^5+16x^4+4x^3

Step-by-step explanation:

A=[2x^3-x+(8x^2+3x)]*4x^2/2=2x^2*(2x^3+8x^2+2x)

or A=4x^5+16x^4+4x^3

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WILL MARK BRAINLIEST IF CORRECT!!!!!!!!!!!!!!!!!!!!

Kipish [7]

Answer:

4000 times

Step-by-step explanation:

The numbers from 1 to 9999 can be shown in the form of:

3xyz, x3yz, xy3z. xyz3

Each of x, y, z represent the digits from 0 to 9, so 10 ways each.

For each of the forms we have:

10*10*10 = 1000 ways and

Total of:

1000*4 = 4000 ways

By the way this is applicable to any digit.

3 0

3 years ago

(CO 1) You need to study the satisfaction of customers of a specific restaurant. You decide to randomly select a few tables and

Elena-2011 [213]

Answer:

Cluster

Step-by-step explanation:

The Cluster sampling technique involves choosing a population to be studied and dividing it into clusters that truly represent the population, through random sampling. In the question above, all the customers in the restaurant represent the population to be studied. Since all the customers can not be interviewed at that time, clusters, in the form of tables are formed through random sampling.

It is necessary for the researcher to ensure that the customers in all the table clusters truly represent the population.

7 0

3 years ago

A polygon has the following coordinates: A(-4,-3), B(4,-3), C(4,-7), D(-4,-7). Find the length of BC.

KIM [24]

Answer:

Step-by-step explanation:

8 0

4 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

Answer ASAP!.......... I need help on 34 more questions.

lianna [129]

Step-by-step explanation:

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3 years ago