Consider the graphs of f(x) = 5 sin (2x-

NEED ANSWER RIGHT NOW PLS Which graph represents the solution set of the inequality Negative 9 greater-than x?

Agata [3.3K]

Answer:

i think C

Step-by-step explanation:

tell me if i am wrong and hope you have a good day

3 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

PLS HELP SHOW ALL YOUR WORKING OUT BRAINLIEST! WILL BE GIVEN :D

aliina [53]

Answer:

(a) Five

(b) Thirteenth

Step-by-step explanation:

We have the mean is 12.6 and

that 11 occurs 4 times, 12 occurs 7 times, 13 occurs 9 times, and 14 occurs f times.

How many children are there in all (that is the sum of the frequencies).

4+7+9+f

20+f

Alright so the mean could be found by doing:

$\frac{4(11)+7(12)+9(13)+f(14)}{20+f}$

But we are given this is also equal to: 12.6.

So we have

$\frac{4(11)+7(12)+9(13)+f(14)}{20+f}=12.6$

I'm going to simplify what I can on top.

$\frac{245+14f}{20+f}=12.6$

I'm going to write 12.6 as $\frac{12.6}{1}$ because I want to cross multiply:

$\frac{245+14f}{20+f}=\frac{12.6}{1}$

$(245+14f)(1)=12.6(20+f)$

Distribute:

$245+14f=252+12.6f$

Subtract 12.6f on both sides:

$245+1.4f=252$

Subtract 245 on both sides:

$1.4f=7$

Divide both sides by 1.4:

$f=5$

There are five 14 years old.

(b) I would say 13 is good age to represent this bunch.

The means was 12.6 which when rounded is 13.

The mode is 13 because it is the most occurring

The median is also 13. Why? If you list out the data 13 will be the middle number. Or you could say there are 25 kids and if I divide it by 2, I get 12.5. This means you only need to count to the 13th kid with the ages in order to tell with the median is.

There are 4 eleven yr olds.

There are 7 twelve yr olds. That is 11 kids so far.

So the median has to be included in the 9 thirteenth yr olds.

3 0

3 years ago

Five men working six hours a day take eight days to fill a trench. How long does it take 3 men working eight hours a day to fill

Oduvanchick [21]

Answer:

The correct answer is 10 days.

Step-by-step explanation:

To fill a trench, 5 men work for 6 hours a day for eight days.

Total work hours required is given by 5 × 6 × 8 = 240 hours.

The same work is supposed to be done by 3 men working 8 hours a day.

Let these three men need to work for x days.

Therefore total work hour these group of three men gave = 3 × 8 × x = 24x hours

Therefore the work hour of both the group of 5 men and 3 men should be equal.

⇒ 24x = 240

⇒ x = 10

Therefore the group of 3 men have to work for 10 days to fill the trench.

6 0

3 years ago

A bakery charges 15 for delivery plus 3 per cupcake. Let C represents the number of cupcakes. Which expression can be used to fi

amid [387]

Answer:

The answer is A.15c+3

Step-by-step explanation:

4 0

2 years ago