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3 years ago

Which term describes the point where three medians of a triangle intersect

Mathematics

2 answers:

Keith_Richards [23]3 years ago

7 0

centriod hope this will help

Hitman42 [59]3 years ago

6 0

The correct answer is centriod

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The cost for 5 dresses for $22 22 dollars $30 35 dollars $36

ipn [44]

<span>The expression:

2-². 2² + 2 (3²) ² + 18 / 8² / ³

it's the same as:

0BS. the number 2 the start of expression is elevated to -2 and the whole expression is being divided by 8 2/3 high.</span>

4 0

3 years ago

Read 2 more answers

What’s the ratio also

dem82 [27]

Answer:

3cm³:20cm³

Step-by-step explanation:

5 * 6 * 20 = 600cm³

4 * 9 * 5 = 180 / 2= 90cm³

90:600

9:60

3:20

7 0

3 years ago

A(n) _____ is any side of a polygon that shares a side with only one angle of a pair of angles.

erastovalidia [21]

Answer:

non-included

Step-by-step explanation:

A non included side of a polygon shares a side with only one angle of a pair of angles.

4 0

3 years ago

Read 2 more answers

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.

3 0

3 years ago

I need help with this one and the others .

timurjin [86]

20 and 0 are your answers

4 0

3 years ago