All categories

2 years ago

Need the answer for 18 please!!!

Mathematics

1 answer:

Sergeu [11.5K]2 years ago

6 0

Answer:

D (2, 1)

Step-by-step explanation:

Fir an in-detailed representation, image below

Started with plotting the given center (yellow)
From the center: go 5 units up, down, right and left (light blue)
You will have 5 points plotted now
connect the light blue dots to form a circle (red, not to scale!)
Now, locate every choice point, which one is on the red line?
D!!!!!

Learn more about Circles here: brainly.com/question/10368742

You might be interested in

The height, h, in feet of a rock above the ground is given by the equation h = -16t^2 + 24t + 16, where t is the time in seconds

boyakko [2]

The rock will hit the ground at 2 seconds.

Explanation:

Given that the height, h, in feet of a rock above the ground is given by the equation $h=-16t^2+24t+16$ , where t is the time in seconds after the rock is thrown.

We need to determine the time at which the rock will hit the ground.

To determine the time, let us equate the height h = 0 in the equation $h=-16t^2+24t+16$ , we get,

$0=-16t^2+24t+16$

Switch sides, we have,

$-16 t^{2}+24 t+16=0$

Let us solve the equation using the quadratic formula.

Thus, we have,

$t=\frac{-24 \pm \sqrt{24^{2}-4(-16) 16}}{2(-16)}$

Simplifying, we get,

$t=\frac{-24 \pm \sqrt{576+1024}}{-32}$

$t=\frac{-24 \pm \sqrt{1600}}{-32}$

$t=\frac{-24 \pm 40}{-32}$

Hence, the two values of t are

$t=\frac{-24 + 40}{-32}$ and $t=\frac{-24 -40}{-32}$

Simplifying the values, we get,

$t=\frac{16}{-32}$ and $t=\frac{-64}{-32}$

Dividing, we get,

$t=-\frac{1}{2}$ and $t=2$

Since, t cannot negative values, we have,

$t=2$

Thus, the rock will hit the ground at 2 seconds.

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:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$