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Vadim26 [7]
4 years ago
7

Which graph represents a function

Mathematics
1 answer:
FromTheMoon [43]4 years ago
4 0

Answer:

Not really an answer just warning you didn't actually attach a graph

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According to einstein, nothing in the universe can move faster than which approximate speed? 1.0 × 107 m/s 1.0 × 108 m/s 3.0 × 1
Helga [31]

The speed of light is 3.0 ×108m/s.

We have ask about the theory of Albert Einstein,

According to einstein, nothing in the universe can move faster than light

<h3>What is the speed of the light?</h3>

Albert Einstein said that nothing travels faster than the speed of light, and the speed of light is

3.0\times108 m/s

So that the nothing can travels faster than the speed of light.

light travels with very high speed.

Therefore we get the correct option is 3.

That is the speed of light is 3.0 ×108m/s.

To learn more about the speed of light visit:

brainly.com/question/104425

4 0
2 years ago
50 Points! - Pls Help!<br><br> TY!
Contact [7]

Answer:

Whats the choices

Step-by-step explanation:

Btw its only 5 pts

3 0
3 years ago
If X is a binomial random variable associated with a binomial experiment with
Makovka662 [10]
For a binomial distribution over n trials and with success probability p, the mean \mu and standard deviation \sigma ar such that

\begin{cases}\mu=np\\\sigma=\sqrt{np(1-p)}\end{cases}

6=\sqrt{np(1-p)}\implies np-np^2=36
54=np\implies54-54p=36\implies p=\dfrac13
54=\dfrac n3\implies n=162
4 0
3 years ago
The perimeter is 54 m. The apothem is \root(3)(3) . To the nearest tenth, what is the area of this regular polygon?
Serjik [45]

The area of this regular polygon is 46.8 square units

<h3>How to determine the area of this regular polygon?</h3>

The given parameters are:

Perimeter, p = 54 m

Apothem, a = √3

The area of this regular polygon is then calculated as:

Area = 0.5 * Apothem * Perimeter

Substitute the known values in the above equation

Area = 0.5 * √3 *  54

Evaluate the product

Area = 46.8

Hence, the area of this regular polygon is 46.8 square units

Read more about area at:

brainly.com/question/14137384

#SPJ1

3 0
2 years ago
Evaluate the surface integral. S xz dS S is the boundary of the region enclosed by the cylinder y2 + z2 = 16 and the planes x =
bagirrra123 [75]

If you project S onto the (x,y)-plane, it casts a "shadow" corresponding to the trapezoidal region

T = {(x,y) : 0 ≤ x ≤ 10 - y and -4 ≤ y ≤ 4}

Let z = f(x, y) = √(16 - y²) and z = g(x, y) = -√(16 - y²), each referring to one half of the cylinder to either side of the plane z = 0.

The surface element for the "positive" half is

dS = √(1 + (∂f/∂x)² + (∂f/dy)²) dx dy

dS = √(1 + 0 + 4y²/(16 - y²)) dx dy

dS = √((16 + 3y²)/(16 - y²)) dx dy

The the surface integral along this half is

\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16-y^2} \sqrt{\frac{16+3y^2}{16-y^2}} \, dx \, dy

\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16+3y^2}\, dx \, dy

\displaystyle \iint_T xz \,dS = \frac12 \int_{-4}^4 (10-y)^2 \sqrt{16+3y^2} \, dy

\displaystyle \iint_T xz \,dS = 416\pi

You'll find that the integral over the "negative" half has the same value, but multiplied by -1. Then the overall surface integral is 0.

8 0
3 years ago
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