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3 years ago

Just question part d) pls

Mathematics

1 answer:

Vinil7 [7]3 years ago

5 0

Answer:

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Step-by-step explanation:

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I don’t get this one but the question was something about finding the equation.

Alina [70]

9514 1404 393

Answer:

y = -(x -3)^2 +2

Step-by-step explanation:

The vertex form of the equation for a parabola is ...

y = a(x -h)^2 +k

where the vertex is (h, k) and the value 'a' is a vertical scale factor.

The value of 'a' can be found by looking at the y-value of points ±1 either side of the vertex relative to the vertex. Here, the vertex y-value is +2 at x=3, and either side goes down 1 unit (to y=1) for 1 unit to the right or left. So, a = -1.

Using the values we've read from the graph for the vertex (h, k) = (3, 2) and the scale factor a = -1, we can write the equation as ...

y = -(x -3)^2 +2

4 0

3 years ago

Tell whether the number is prime or composite 68

Readme [11.4K]

68 is a composite number

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3 years ago

JOIN <br> id 716 2342 9565<br><br> pass Nike4

pickupchik [31]

Answer: sure

Step-by-step explanation:

7 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

Express in the form <br> 1<br> :<br> n<br> .<br> Give <br> n<br> as a decimal.<br> 10<br> :<br> 4

Brut [27]

Answer:

n=0.4

Step-by-step explanation:

10:4=5:2

to make 5 as 1, we will divide both 5 and 2 by 5

<h3> $\frac{5}{5}$ : $\frac{2}{5}$ =1:n</h3><h3>therefore, n= $\frac{2}{5}$ = 0.4</h3><h3 /><h3> </h3>

5 0

3 years ago