(49p2–490p) ÷(p–10) please help!!
1 answer:
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Answer:
Step-by-step explanation:
We have the following expression:
We can find the GCF with the following steps.
1. Find the GCF for the numerical part 15 and -21
The factors for 15 are : 1,3,5,15
The factors for -21 are -21,-7,-3,-1,1,3,7,21
And the common factors are 1,3
The GCF for the numerical part is 3*1 = 3
2. Find the GCF for the variable
The factors of x^2 are : x*x
The factors of x are: x
The common factors are x for this case, so then the GCF for the variable part is x
3. Multiply the two results from steps 1 and 2
GCf= 3*x = 3x
If we consider the new expression:
We can find the GCF with the following steps.
1. Find the GCF for the numerical part 15 and -21
The factors for 15 are : 1,3,5,15
The factors for -21 are 1,3,7,21
And the common factors are 1,3
The GCF for the numerical part is 3*1 = 3
2. Find the GCF for the variable
The factors of x^2 are : x*x
The factors of x are: x
The common factors are x for this case, so then the GCF for the variable part is x
3. Multiply the two results from steps 1 and 2
GCf= 3*x = 3x
And as we can see both GCF are equal. So then we satisfy the condition required.
Answer:
3
Step-by-step explanation:
Given that:
The number of paint sets, Suzanne bought= 6
As each set contains the same number of brushes.
Let x be the number of brushes each set contains.
So, the total number of brushes in 6 sets = 6x.
It is given that the total number of brushes she has is 18, which is equals to 6x.
So, the required equation is
Hence, the number of paint brushes each set contains is 3.