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1 year ago

Find the equation of the plane that goes through three points:

Mathematics

1 answer:

frosja888 [35]1 year ago

7 0

The equation of the plane that goes through these points is:

6x + 2y + z = 10.

<h3>How to find the equation of a plane given three points?</h3>

The equation of the plane is found replacing the points into the following equation:

ax + by + c = z.

For point A, we have that:

3b + c = 4.

For point B, we have that:

a + 2b + c = 0.

For point C, we have that:

-a + 6b + c = 4.

Hence the system is:

3b + c = 4.

a + 2b + c = 0.

-a + 6b + c = 4.

From the first equation, we have that:

c = 4 - 3b.

Replacing in the second, we have that:

a + 2b + 4 - 3b = 0

a - b = -4.

Replacing in the third, we have that:

-a + 6b + 4 - 3b = 4.

-a + 3b = 0.

a = 3b.

We have that a - b = -4, hence:

3b - b = -4

2b = -4

b = -2.

a = 3b, hence a = -6.

c = 4 - 3b -> c = 10.

Hence the equation is:

ax + by + c = z.

z = -6x - 2y + 10

6x + 2y + z = 10.

More can be learned about the equation of a plane at brainly.com/question/13854649

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Judys brother sam has a collection of 96 comic book. What are the ten way sam could divide his comic books into equal groups?

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3 years ago

DIscrete Math

Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

$f : X \rightarrow Y$ is surjective implies there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ .
If there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ , then $f : X \rightarrow Y$ is surjective

Let us start by the first implication.

Our hypothesis is that the function $f : X \rightarrow Y$ is surjective. From this we know that for every $y\in Y$ there exist, at least, one $x\in X$ such that $y=f(x)$ .

Now, define the sets $X_y = \{x\in X: y=f(x)\}$ . Notice that the set $X_y$ is the pre-image of the element $y$ . Also, from the fact that $f$ is a function we deduce that $X_{y_1}\cap X_{y_2}=\emptyset$ , and because $f$ the sets $X_y$ are no empty.

From each set $X_y$ choose only one element $x_y$ , and notice that $f(x_y)=y$ .

So, we can define the function $h:Y\rightarrow X$ as $h(y)=x_y$ . It is no difficult to conclude that $f\circ h(y) = f(x_y)=y$ . With this we have that $f\circ h=1_Y$ , and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function $h:Y\rightarrow X$ such that $f\circ h=1_Y$ .

Take an element $y\in Y$ , then $f\circ h(y)=y$ . Now, write $x=h(y)$ and notice that $x\in X$ . Also, with this we have that $f(x)=y$ .

So, for every element $y\in Y$ we have found that an element $x\in X$ (recall that $x=h(y)$ ) such that $y=f(x)$ , which is equivalent to the fact that $f$ is surjective. Therefore, the prove is complete.

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Ksivusya [100]

Answer: 67

Step-by-step explanation:

3x^2 + 5x + 25

x = 3

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3^2 = 9

3(9) + 5(3) + 25

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