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suter [353]
3 years ago
5

The polynomial given below has _____ root(s).

Mathematics
1 answer:
lisov135 [29]3 years ago
8 0
3x²-8x+4=0
(3x-2) (x-2)=0
x=2/3 ,  x = 2
Hence, the polynomial has two positive roots.
Option A is the correct choice.
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Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y
DiKsa [7]

The area of the surface is given exactly by the integral,

\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx

We have

y(x)=\dfrac15x^5\implies y'(x)=x^4

so the area is

\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the i-th subinterval are, respectively,

\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2

r_i=\dfrac{5-0}{10}i=\dfrac i2

with midpoint

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

with 1\le i\le10.

Over each subinterval, we interpolate f(x)=\sqrt{1+x^8} with the quadratic polynomial,

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

Then

\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that the latter integral reduces significantly to

\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)

which is about 651.918, so that the area is approximately 651.918\pi\approx\boxed{2048}.

Compare this to actual value of the integral, which is closer to 1967.

4 0
3 years ago
Can a Right Triangle be a(n)<br> A.Obtuse<br> B.Scalene
Blizzard [7]
Your answer is B hope i helped
4 0
3 years ago
Read 2 more answers
Y = 4x-9<br> y = x-3<br> Which method and why? Solve, shoving all work
vodomira [7]

Answer:

(2,-1)

Step-by-step explanation:

The ys in both questions are isolated on one side of the equation in both. The numerical coefficient of both is 1. Therefore you should equate the the left side of each equation to the other left side. The solution is the easiest one to solve because there is only 1 unknown on both sides.

4x - 9 = x - 3                      Subtract x from both sides.

4x-x - 9 = -3                       Combine

3x -9 = - 3                          Add 9 to both sides

3x - 9+9 = - 3+9

3x = 6                                Divide by 3

3x/3 = 6/3

x = 2

===============

Now use the second equation to solve for y

y = x - 3

y = 2 - 3

y = -1

==============

The solution is (2,-1)

7 0
3 years ago
Suppose that a tunnel is dug through the center of the earth. You stand on a tower 100 m tall and drop a 3-kg rock into the tunn
Mashutka [201]

Answer:

The frequency of the resulting harmonic motion is 0.000219 Hz

Step-by-step explanation:

We are going to calculate the time it takes for one single wave ocillation.

Frequency and the time taken to finish a single wave oscillation are inversely proportional. The formula for calculating frequency when given the time taken to complete a wave cycle is written as: f = 1 / T

In this formula, f represents frequency and T represents the time period or amount of time required to complete a single wave oscillation.

I consider the initial speed to be zero, because it is of no significance compared with the free fall into the earth, through the earth and back again.

Given from Wikipedia:

The diameter of the earth is 1.2742 * 10⁴ km which is 1.27 * 10⁷ m

2 times the radius = diameter, so the radius of the earth = (1.27 * 10⁷ m) /2 = 6.4 * 10⁶ m

radius earth = r

r = 6.4 * 10⁶ m

Now imagine the tunnel and the free fall.

1. Initially the rock has no speed.

2. Due to the gravitational accelleration, the rock will increase it's speed every second by a factor of 9.8.

3. The Rock gains speed untill it reached the centre of the earth. By then it will have reached it's maximum speed and it has travelled the distance r !

4. After this moment, the Rock will be slowed down because of the negative accelleration...

After it has travelled from the centre of the earth to the other end of the earth, it will have stopped completely, and again passing the distance r.

5. Now at the other end of the earth there is the same initial situation as described at point 1, only the Rock has travelled the distance equal to the diameter of the earth, (exactly 2 times r).

So basically, the samething happens once more, only this time it starts exactly from the other end of the earth...

6. Initially the rock has no speed.

7. Due to the gravitational accelleration, the rock will increase it's speed every second by a factor of 9.8.

8. The Rock gains speed untill it reached the centre of the earth. By then it will have reached it's maximum speed.

9. By now the Rock will be slowed down because of the negative accelleration... It is moving towards the initial starting point...

After it has travelled from the centre of the earth to the other end of the earth, it will have stopped completely.

10. Now finally the Rock is exactly at the starting position.

In reality there will have been some loss of speed due to friction, so the Rock will be slightly lower then the 100 m above the ground.

let's calculate the time it takes to free fall for the distance r.

initial speed =0 and after 6.4 * 10⁶ m it's speed will be maximum. We need to find out how much time passes before that distance is passed.

r = v*t + 0.5*a*t²

r = 0 + 0.5*a*t²

0.5*a*t² = r

t² = r / ( 0.5 * a )

t² = 6.4 *10⁶ / ( 0.5 * 9.8 )

t² = 1.306 * 10 ⁶

t = 1142.86 s

Now please confirm that in order for the Rock to move back to the initial starting point it has to travel 4 times as much time. It has to travel r to centre of the earth then another r to travel to to the other side of the earth, and back again. So indeed 4 times r.

The time it will take must be the same as 4 * 1142.86 s

now this is the time of one single wave ocillation.

Since T = 4571.43 s

f = 1 / 4571.43

f = 0.00021874993164 Hz

The frequency of the resulting harmonic motion is 2.19 *10-4

The frequency of the resulting harmonic motion is 0.000219 Hz

6 0
3 years ago
Identify the vertex and the y-intercept of the graph translated 1 unit right from the parent function f(x)=x2
Ronch [10]

Answer:

Step-by-step explanation:

y = (x-1)²

vertex (1,0)

y-intercept = (0-1)² = 1

5 0
3 years ago
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