All categories

3 years ago

Find the greatest common factor of 10, 30,10,30,10, comma, 30, comma and 454545.

Mathematics

1 answer:

borishaifa [10]3 years ago

5 0

Answer:

GCF = 5

Step-by-step explanation:

Given:

The numbers are given as:

10, 30 and 45

A factor of a number is a number by which the given number is evenly divisible.

Example: 2 is a factor of 4 as $4\div 2 =2$ . So, 4 is evenly divisible by 2 and hence 2 is a factor of 4.

Now, factors of 10 = 1, 2, 5, 10.

Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30.

Factors of 45 = 1, 3, 5, 9, 15, 45.

So, the greatest common factor in all the three numbers is 5.

Hence, the greatest common factor of 10, 30 and 45 is 5.

You might be interested in

Kevin cycles 18 miles every morning as part of his exercise. The time it takes him to complete the distance varies inversely as

Schach [20]

Hello, I am papaguy

Here is your answer

In terms of distance, the average speed for the total x miles you'd travel would be 50 miles per hour. However, if you include the fact that you spent less time traveling 60 miles per hour than 40, the average would be closer to 40 than 60 miles per hour.

Please note this is information you can use do not copy and paste.

Thank you papaguy And If I am wrong I will look into it!

8 0

3 years ago

Read 2 more answers

What IS the aswer PLZZZZZZZ help

BabaBlast [244]

Answer:

ONE PAW AND MAW

Step-by-step explanation:

5 0

2 years ago

A. For every 6 hours she works, her earrings go up by $1.

grandymaker [24]

Answer:

the answer is c

Step-by-step explanation:

as it shows on the graph every hour the earrings increase by 6$

6 0

3 years ago

A manufacturing company has two retail outlets. It is known that 30% of all potential customers buy

aleksandrvk [35]

Answer:

(a) A y P(A) = 0.4 (b) $\bar{B}$ y $P(\bar{B})$ =0.5 (c) $\bar{A}$ ∪ $\bar{B}$ y P( $\bar{A}$ ∪ $\bar{B}$ ) = 0.9 (d) $\bar{A}$ ∩ $\bar{B}$ y P( $\bar{A}$ ∩ $\bar{B}$ )=0.2

Step-by-step explanation:

A was defined as the event that a potential customer, randomly chosen, buys from outlet 1 in the original problem statement. We know that B denotes the event that a randomly chosen customer buys from outlet 2. So

P( $A\cap \bar{B}$ ) = 0.3, P( $B\cap \bar{A}$ ) = 0.4 and P( $A\cap B$ ) = 0.1

(a) P(A) = P( $A\cap (B\cup\bar{B})$ ) = P( $A\cap B$ ) + P( $A\cap \bar{B}$ ) = 0.1 + 0.3 = 0.4

(b) P(B) = P( $B\cap (A\cup\bar{A})$ ) = P( $B\cap A$ ) + P( $B\cap \bar{A}$ ) = 0.1 + 0.4 = 0.5

P( $\bar{B}$ ) = 1-P(B) = 1-0.5 = 0.5

(c) The customer does not buy from outlet 1 is the complement of A, i.e., $\bar{A}$ , and the customer does not buy from outlet 2 is the complement of B, i.e., $\bar{B}$ , so, the customer does not buy from outlet 1 or does not buy from outlet 2 is $\bar{A}$ ∪ $\bar{B}$ and P( $\bar{A}$ ∪ $\bar{B}$ ) = P( $(A\cap B)^{c}$ ) by De Morgan's laws

P( $(A\cap B)^{c}$ ) = 1-P(A∩B)=1-0.1=0.9

(d) The customer does not buy from outlet 1 is the complement of A, and the customer does not buy from outlet 2 is the complement of B, so we have that the statement in (d) is equivalent to $\bar{A}$ ∩ $\bar{B}$ and P( $\bar{A}$ ∩ $\bar{B}$ ) = P( $(AUB)^{c}$ ) by De Morgan's laws, and