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ziro4ka [17]
3 years ago
11

The (greatest, least) of the common factors of 2 or more numbers is the (greatest, least) common factor of the numbers.

Mathematics
1 answer:
Liula [17]3 years ago
4 0

Answer:

The greatest of the common factors of 2 or more numbers is the greatest common factor of the numbers.

Step-by-step explanation:

As we know that

The greatest common factor of two or more whole numbers is the largest whole number that divides evenly into each of the numbers.

Thus, the greatest of the common factors of 2 or more numbers is the greatest common factor of the numbers.

For example, lets find the greatest common factor of 36 and 54.

The possible factors of 36 are:

                                                   1, 2, 3, 4, 6, 9, 12, 18, and 36.

The possible factors of 54 are:

                                                   1, 2, 3, 6, 9, 18, 27, and 54.

The common factors of 36 and 54 are:

                                                   1, 2, 3, 6, 9, 18

So, from the common factors, it is easy to figure out that 18 is the greatest common factor.

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-3 + -2 = -5

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Biologists have found that the number of chirps some types of cricket make per minute depends upon the temperature this relation
Alex Ar [27]

Answer: Crickets make 172 chirps/min at 80 degrees Fahrenheit

Step-by-step explanation:

Step 1

Using the point slope formulae for Linear functions , we have that

y - y1 = m(x - x1)

where x  and y are two points  representing the  temperature and chirps /min respectively

And  

m= slope

Step 2 : Finding the slope , m

where x1=60

           y1=92

           x2= 75

           y2=152

m= (y2-y1)/ (x2-x1)

= (152- 92)/(75-60)

=60/15 =4

Bringing down the point/slope formula and imputing the known values to find our equation to show the relationship cricket make per minute and  the temperature

y - y1 = m(x - x1)

y - 92= 4(x - 60)

y - 92 = 4x -240

y = 4x - 240 + 92

y = 4x - 148

Step 3

To find the number of chirps be per minute (y) if the temperature is 80 degrees Fahrenheit( x)

y = 4(80) - 160

y = 320 - 148

y = 172 chirps/min at 80 degrees Fahrenheit

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2 years ago
In need of help PLEASE!!!!​
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Answer:

a bc I think it's right idk really

Step-by-step explanation:

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Solve the following systems by substitution
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Answer:

The answer to your question is below

Step-by-step explanation:

Question 1

                x = 5                       Equation l

                2x + y = 10              Equation ll

- Substitute Equation l in equation ll

               2(5) + y = 10

                y = 10 - 10

                y = 0

- Solution  (5, 0)

Question 2

                 x + 16y = 20          Equation l

                 x = 4y                    Equation ll

Substitute equation ll in equation l

                4y + 16y = 20

                        20y = 20

                           y = 20/20

                           y = 1

-Find x

                x = 4(1)

                x = 4

-Solution

                (4, 1)

Question 3

                   2x + 8y = 20            Equation l

                     x = 2                       Equation ll

-Substitute equation ll in equation l

                   2(2) + 8y = 20

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                            8y = 20 - 4

                            8y = 16

                              y = 16/8

                              y = 2

- Solution

                   (2, 2)              

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Answer:

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Enlargement

Step-1

Let's dilate triangle PQR by a scale factor of 2 with a center of dilation at point C.

Step-2

First, we will use a straightedge to connect the center of dilation C to each vertex. We are going to extend those lines across the whole page.

Step-3

Next, we'll place the point of the compass on the center of dilation C and the pencil on vertex Q and measure that distance.

Step-4

Now, without changing the size of the compass, we will move the point of the compass to vertex Q, and make a mark on the line that is extended through Q.

Step-5

Notice that since we measured the distance from C to Q and then used that same distance to mark the line, we have now created a new point that is twice the distance from C that Q was.

We now repeat the process for each of the other vertices of the triangle.

Step-6

These marks represent the vertices of our dilated image. We often use the same letters with an apostrophe to show that it is the corresponding vertex.

Step-7

Finally, we use the straightedge to connect all of the new vertices.

Now,See how triangle P

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Q

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R

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is twice the size of the original triangle.

Reduction

Step-1

Given triangle PQR, now let's dilate the image with a center of dilation at C and with a scale factor of

2

1

Step-2

Using the straightedge, we will connect all of the vertices to the center of dilation C. In this case, since our scale factor is less than 1, we are reducing the size of the triangle.

Step-3

The green lines are the new lines that we have drawn.

Since we have to reduce the size of the triangle by half, we want to cut the distance in half from each vertex to the center of dilation. In order to do this, we will set our compass to a distance that is more than halfway between vertex P and point C but less than the total distance.

Step-4

Keeping the compass at that same distance, start with the point of the compass on vertex P, then draw a curve that intersects the line between P and C. Next, keeping the compass at the same distance, place the point of the compass on C and draw a curve that intersects the line between P and C and intersects the previously drawn curve.

Step-5

Now, we will take the straightedge and connect the points where the curves overlap.

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