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4 years ago

Which geometric figure consists of the set of all points in a plane that are the same distance from a given point?

2 answers:

6 0

An arc, a circle and a sphere describe a geometric figure consisting of the set of points in a plane that are the same distance from a given point.

An arc is any two connected sections of the circumference of a circle. Whereas an arc consists of the set of points in a plane that are the same distance from a given point it does not consist ALL the points.

A circle is a simple closed shape in consisting of the set of all points in a plane that are at a given distance from a given point, called the centre of the circle. A circle is a 2-dimensional figure and thus does not contain set of points in the 3-dimensional plane. Hence a circle does not consist ALL the points in a plane that are at a given distance from a given point.

A sphere is a perfectly round geometrical object in three-dimensional space that is the surface of a completely round ball. Like a circle, which geometrically is an object in two-dimensional space, a sphere is defined mathematically as the set of points that are all at the same distance, r, from a given point, but in three-dimensional space. This distance, r, is the radius of the sphere, and the given point is the center of the sphere.

Thus, the sphere consists of the set of ALL points in a plane that are the same distance from a given point.

CaHeK987 [17]4 years ago

3 0

Answer:

An arc, a circle and a sphere describe a geometric figure consisting of the set of points in a plane that are the same distance from a given point.

An arc is any two connected sections of the circumference of a circle. Whereas an arc consists of the set of points in a plane that are the same distance from a given point it does not consist ALL the points.

Thus, the sphere consists of the set of ALL points in a plane that are the same distance from a given point.

Step-by-step explanation:

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In the general population, the mean IQ is 100 with a standard deviation of 15. Dr. Robbins wants to see if a class of statistics

Shalnov [3]

From the information given, a right-tailed test using the z-distribution is appropriate.

At the null hypothesis, it is tested if the class of statistics students is not significantly smarter than the general population, that is, their mean IQ is of at most 100, hence:

$H_0: \mu \leq 100$

At the alternative hypothesis, it is tested if they are significantly smarter, that is, their mean IQ is greater than 100, hence:

$H_1: \mu > 100$

We are testing if the mean is greater than a value, hence it is a right-tailed test.

We have the standard deviation for the population, hence, the z-distribution is used.

A similar problem is given at brainly.com/question/23413408

7 0

2 years ago

Create a polynomial that is added to 5x2 + 6x + 9 to get 7x2 + 9x + 14.

DiKsa [7]

Answer:

Step-by-step explanation:

Subtract first from second

= 2x^2 + 3x + 5

4 0

3 years ago

A red die, a blue die, and a yellow die (all six sided) are rolled. we are interested in the probability that the number appeari

NISA [10]

5/54 or approximately 0.092592593
There are 6^3 = 216 possible outcomes of rolling these 3 dice. Let's count the number of possible rolls that meet the criteria b < y < r, manually.
r = 1 or 2 is obviously impossible. So let's look at r = 3 through 6.
r = 3, y = 2, b = 1 is the only possibility for r=3. So n = 1
r = 4, y = 3, b = {1,2}, so n = 1 + 2 = 3
r = 4, y = 2, b = 1, so n = 3 + 1 = 4
r = 5, y = 4, b = {1,2,3}, so n = 4 + 3 = 7
r = 5, y = 3, b = {1,2}, so n = 7 + 2 = 9
r = 5, y = 2, b = 1, so n = 9 + 1 = 10

And I see a pattern, for the most restrictive r, there is 1 possibility. For the next most restrictive, there's 2+1 = 3 possibilities. Then the next one is 3+2+1
= 6 possibilities. So for r = 6, there should be 4+3+2+1 = 10 possibilities.
Let's see
r = 6, y = 5, b = {4,3,2,1}, so n = 10 + 4 = 14
r = 6, y = 4, b = {3,2,1}, so n = 14 + 3 = 17
r = 6, y = 3, b = {2,1}, so n = 17 + 2 = 19
r = 6, y = 2, b = 1, so n = 19 + 1 = 20
And the pattern holds. So there are 20 possible rolls that meet the desired criteria out of 216 possible rolls. So 20/216 = 5/54.

6 0

3 years ago

Scrat [10]

If two similar triangles have side lengths 3cm and 7cm then we can determine that the scale factor for the lengths is: 7/3

Now, we have to work out the area scale factor !

This is simple, since area is given in cm² then all we have to do is square our original scale factor!

So,

7/3² = 49/9

So, we can say that the ratio of the area would be:

x : 49/9x

Hope this helps! <3

8 0

3 years ago

Brayden, Howard, and Vincent are on the football team. During a game, Brayden gains 1 less than 5 times the average number of ya

kotegsom [21]

Answer: They are parallel so there is no solution.

Step-by-step explanation:

Brayden- y=5x-1 or y-5x= -1

Howard- y=5x+1 or y-5x= 1

Vincent- y=4x+5 or y-4x= 5

In summary, they are all parallel so there is no solution.

3 0

3 years ago

Read 2 more answers