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3 years ago

The Neolithic Revolution

Mathematics

2 answers:

Amanda [17]3 years ago

6 0

The Neolithic Revolution allowed people to live in permanent settlements.

Explanation:

The Neolithic Revolution is the phase of transition in which many of the human cultures in the world shifted their lifestyle to permanent settlement and agriculture from hunting and gathering.

This era is also named as First Agricultural Revolution or Neolithic Demographic Transition. The agricultural practices led the cultures to develop the knowledge of domestication of animals and they began to observe and experiment with the plants (to learn or know how they grow and develop).

Neolithic period is named as revolution because this period changed the way of life of most of the communities. This first occurred in so-called "Fertile Crescent" or Mesopotamia (modern Iraq).

shutvik [7]3 years ago

6 0

The Neolithic Revolution allowed people to live in permanent settlements.

Answer: Option B

Step-by-step explanation:

This phase 'Neolithic Revolution' witnessed the change in the behavior of the people, notably the development of habits like hunting and gathering of animals and fruits.

Although it is a development of habits, people refer to it as a revolution because it changed the lifestyle of the communities.
It dates back to prehistoric times which is said to have happened almost nine thousand years ago. Thereby concluding that, the Neolithic revolution allowed people to live in permanent settlements.

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SashulF [63]

The Solution.

The given equation is

$\begin{gathered} y=|x|+7 \\ \text{Where x=11} \end{gathered}$

Substituting 11 for x, we can get the value of y.

$\begin{gathered} y=|11|+7 \\ y=11+7 \\ y=18 \end{gathered}$

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At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not

Nikitich [7]

If $f(x, y, z) = c$ represent a family of surfaces for different values of the constant $c$ . The gradient of the function $f$ defined as $\nabla f$ is a vector normal to the surface $f(x, y, z) = c$ .

Given the paraboloid

$y = x^2 + z^2$ .

We can rewrite it as a scalar value function f as follows:

$f(x,y,z)=x^2-y+z^2=0$

The normal to the paraboloid at any point is given by:

$\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\ \\ =2xi-j+2zk$

Also, the normal to the given plane $3x + 2y + 7z = 2$ is given by:

$3i+2j+7k$

Equating the two normal vectors, we have:

$2x=3\Rightarrow x= \frac{3}{2} \\ \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}$

Since, -1 = 2 is not possible, therefore there exist no such point on the paraboloid $y = x^2 + z^2$ such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2.

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vladimir1956 [14]

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Help in your answer!

mart [117]

Answer:

Two different transformations happen. Rotation/Reflection and translation (I say rotation/reflection because they can both put the parabola where it is after being transformed)

(Rotation- 180 degrees)

(Reflection across its it's starting point)

Translation- pushed up 4 points from negative 4 to 0 on the y-axis

When you put the parent function into a graph it is positive so the u is opening upward and starts at -4. Both sides of the u go through the points (0,0)and (4,0).

But in the transformed function is negative so the u opens downward. The sides going through the points (0,-4) and (4,-4)

I hope this makes sense

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