All categories

3 years ago

50, 40 What’s the GCF of these numbers

Mathematics

2 answers:

ella [17]3 years ago

5 0

Answer: 10

Step-by-step explanation: To find the greatest common factor or <em>GCF</em> of 50 and 40, we begin by finding all of the factors of each number.

To find the factors of 50, we know that 50 ÷ 1 is 50 so 1 and 50 are factors.

50 ÷ 2 is 25 so 2 and 25 are factors.

We can't divide 50 evenly by 3 or 4 so we know they aren't factors.

50 ÷ 5 = 10 so 5 and 10 are factors.

However, if we continue to divide 50 by 6, 7, 8, and so on, we won't find any new factors. So the factors of 50 are 1, 2, 5, 10, 25, and 50.

Now let's find the factors of 40.

40 ÷ 1 is 40 so 1 and 40 are factors.

40 ÷ 2 is 20 so 2 and 20 are factors.

We can't divide 40 evenly by 3.

40 ÷ 4 is 10 so 4 and 10 are factors.

40 ÷ 5 is 8 so 5 and 8 are factors.

However, if we continue to divide by 6, 7, 8, and so on, we won't find any new factors.

So the factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40

Now that we have our list of factors, to find the greatest common factor, we simply find the largest number that is shared by the two lists.

Factors of 50 - 1, 2, 5, 10, 25, and 50

Factors of 40 - 1, 2, 4, 5, 8, 10, 20, and 40

Notice that both lists have a 50 as a factor and there is nothing larger than 50 that appears in both lists. So the greatest common factor of 50 and 40 is 10.

dexar [7]3 years ago

3 0

Answer:

Step-by-step explanation:

40=2 x 2 x 2 x 5

50=2 x 5 x 5

therefore, GFC = 2x5

GFC= 10

You might be interested in

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$