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grigory [225]
3 years ago
11

What is the realationship between the volume of a cone and the volume of a sphere

Mathematics
1 answer:
patriot [66]3 years ago
6 0
Depending on the size, the cone would have more volume
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A line passes through the two given points. Is it vertical, horizontal, or neither?
Leviafan [203]

<u>Answer:</u>

A line passes through the two given points (5,2), (5,-2). The line is vertical line

<u>Solution:</u>

Given, two points are (5, 2) and (5, -2)

We have to find that a line that passes through the given two points is vertical or horizontal or neither.

First let us find the slope of the line that passes through given two points.

So, slope of line m =\frac{y_{2}-y_{1}}{x_{2}-x_{2}} \text { where, }\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right) \text { are two points on line. }

So slope of our line =\frac{-2-2}{5-5}=\frac{-4}{0} = undefined

Here slope of our line is undefined which means it is parallel to the y – axis.

Hence, line passing through given points is vertical line

6 0
3 years ago
HELP PLEASE!!!!!!!! I GIVE 21 POINTS!!!!!!!!! AND BRAINLEAST &lt;3
REY [17]

6. domain=age

range= number of pets

it is not a function because an x value repeats


7. domain=1,2,3,4,5

range= 50, 100, 150, 150, 200

it is a function because the lines don't connect

7 0
3 years ago
HELPPPPPP WILL GIVE BRAINLEST AND THANKS EASY PLEASE HELP
Ipatiy [6.2K]
It is none of them since -2+5 is 3. Are you sure that's the right equation?
7 0
3 years ago
8. Let R be the relation on the set of all sets of real numbers such that SRT if and only if S and T have the same cardinality.
sergejj [24]

Answer:

We must prove that the relation is reflexive, symmetric and transitive. Recall that to sets have the same cardinality if there exist a bijective mapping between them.

<em>Reflexive: </em>Take the identity map I:S\rightarrow S, which is bijective. Then SRS.

<em>Symmetric:</em> If SRT then, there exist a bijective map f:S\rightarrow R. In order to prove that TRS just take the inverse map of f: f^{-1} which is also bijective. Therefore, TRS.

<em>Transitivity: </em>Suppose that SRT and TRU. Also, assume that f is the bijective map between S and T, and g the bijective map between T and U. It is not difficult to check that the map h=g(f) is bijective and h:S\rightarrow U. Therefore, SRU.

Hence, the relation R is an equivalence relation.

The equivalence class of the set {0,1,2} is the class of all the sets with three elements, and we can associate it with the number 3. There is a construction of natural numbers based on this idea.

The equivalence class of Z is the same equivalence class of N. Therefore, is the class of all denumerable or countable sets.

Step-by-step explanation:

When we want to prove that a given relation R is equivalence, we need to check that R satisfies all the three conditions: reflexive, symmetric and transitivity. Usually the first two are very simple to prove and comes directly from the definition. The transitivity is more tricky. In this case we need to recall the definition of cardinality.

7 0
3 years ago
How can you tell if a number is divisible by 3? 
Bingel [31]
The correct answer is B. Try it out with the following number: 312, 1893, 255, 344, 232. Two of the or not divisible by 3 by the way.
3 0
2 years ago
Read 2 more answers
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