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8 months ago

PLEASE HELP VERY URGENT WILL GIVE BRAINILY PLEASE I BEG YOU PLEASE

Mathematics

1 answer:

yuradex [85]8 months ago

5 0

Answer: 20.5 units

Step-by-step explanation:

P=Perimeter

JI=3-(-3)

JI=3+3

JI=6

IK=7-1

IK=6

JK=(6^2+6^2)^1/2 ==> Distance Formula

JK=(36+36)^1/2

JK=(2*36)^1/2

JK=6(2)^1/2

P=JI+IK+JK

P=6+6+6(2)^1/2

P=12+6(2)^1/2

P=20.485 units

P=20.5 units

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Solution:

Given that, we have to find the greatest common factor of $44c^5, 22c^3 , 11c^4$

The greatest number that is a factor of two (or more) other numbers.

When we find all the factors of two or more numbers, and some factors are the same ("common"), then the largest of those common factors is the Greatest Common Factor

$\underline{\text{ GCF of } 44c^5, 22c^3 , 11c^4 :}$

First find the GCF of numbers and then find the GCF of variables and then multiply them together

Step 1: GCF of 44, 22 and 11

The factors of 11 are: 1, 11

The factors of 22 are: 1, 2, 11, 22

The factors of 44 are: 1, 2, 4, 11, 22, 44

11 is the greatest factor that is common in above three list

Then the greatest common factor is 11

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$c^5 = c^3 \times c^2\\\\c^3 = c^2 \times c^1 \text{ or } c^3\\\\c^4 = c^3 \times c^1$

Thus the greatest common factor is $c^3$

Step 3: Multiply the GCF of 44, 22, 11 and GCF of $c^5, c^3 , c^4$

$\text{ GCF of } 44c^5, 22c^3 , 11c^4 = 11 \times c^3 = 11c^3$

Thus $\text{ GCF of } 44c^5, 22c^3 , 11c^4 \text{ is } 11c^3$

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