Aiden wants to find the mass of a bowling ball. Which unit should he use?

Answers

Answer 1
Answer: Kg, but if you use grams just convert it over
Answer 2
Answer: Mass is measured in grams.  he should use grams or kilograms.

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Greatest common factor of 21, 35, 49

Answers

The greatest common factor of 21, 35, and 49 is 49, which is determined by the prime factorization.

To find the greatest common factor (GCF) of 21, 35, and 49, we can start by finding the prime factorization of each number.

Here are the prime factorizations of the given numbers:

  • 21 = 3 x 7
  • 35 = 5 x 7
  • 49 = 7 x 7

To find the GCF, we need to identify the common prime factors and multiply them together.

Here, the only common prime factor is 7, which appears twice in the prime factorization of 49.

Therefore, the GCF of 21, 35, and 49 is 7 x 7 = 49.

So, the greatest common factor of 21, 35, and 49 is 49.

Learn more about the Greatest Common Factor here:

brainly.com/question/18905667

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I think it is 7 because 7 x 3 = 21
7 x 5 = 35
7 x 7 = 49

Select Is a Function or Is not a Function to correctly classify each relation. Title Is a Function Is not a Function {(2, 2),(4, 4),(6, 6),(8, 8)}
{(0, 3),(3, 5),(5, 6),(8, 4)}
{(1, 2),(3, 3),(4, 8),(6, 3)}
{(3, 4),(5, 2),(5, 6),(7, 3)}

Answers

A mapping is said to be function if each element in the domain is related with only one element in the range.Or you can say that a relation is a function also when two different elements in domain have same image or related with same element of the Co domain.

The elements which is related with elements of domain is called Co domain.

So , using the definition of function that i have written above you can classify that the relation is a function or not

1. {(2, 2),(4, 4),(6, 6),(8, 8)}

→→Each element is related with itself , no two different images have different Preimage, For example (2,3),(2,4) that is not happening in this case. so this relation is a function.This kind of function is called identity function.

2. (0, 3),(3, 5),(5, 6),(8, 4)→Different element have different images .So this Relation is a function.


3. (1, 2),(3, 3),(4, 8),(6, 3) →→→Here two elements have same image 3. But still this is a function because two elements in domain or preimage can have same image in a function.


4. (3, 4),(5, 2),(5, 6),(7, 3)→→→This is not a function because two different elements of image have same preimage that is image of 2,and 6 have same preimage which is 5. So this is not a function.

From The given options 1,2,3 are functions.


{(2, 2),(4, 4),(6, 6),(8, 8)} Is a function.
{(0, 3),(3, 5),(5, 6),(8, 4)} Is a function.
{(1, 2),(3, 3),(4, 8),(6, 3)} Is a function.
{(3, 4),(5, 2),(5, 6),(7, 3)} Is not a function. 

For a relation to be a function, every x value must have only one y value.

Hope this helps.

PLEASE HELP I GIVE THANKS

Answers

The range is the difference between the highest and the lowest number.
The highest is 48.
The lowest is 40.

The range is 48-40=8 \Rightarrow C


I think option C) is the correct answer.


Is and angle made of any two rays

Answers

two rays that share a vertex.
If to rays share a vertex than yes

Ben can type 150 words in 3 minutes. how many words can he type in 2minutes

Answers

Ben can type 100 words in 2 minutes. Since 50 can go into 150 three times, you do 50 x 2 which is 100 words in two minutes.
the answer would be 100.
this is because ben is most likely to type words in an equal amount of time. 
you can solve this by first finding how many words ben can type in ONE minute. this results in division or 150 divided by 3 which gives 50.
since the question is asking about TWO minutes, you add 50 two times or multiply 50 by 2 which will either way give you 100


Solve 2x^2 +16x+50=0

Answers

2x^2 +16x+50=0\ \ \ \ \ |:2\n\nx^2+8x+25=0\n\nx^2+2\cdot x\cdot4+25=0\ \ \ \ |+4^2\n\n\underbrace{x^2+2\cdot x\cdot4+4^2}_((a+b)^2=a^2+2ab+b^2)+25=4^2\ \ \ \ \ |-25\n\n(x+4)^2=16-25\n\n(x+4)^2=-9 < 0-NO\ REAL\ SOLUTION

COMPLEX\ SOLUTIONS\n\n(x+4)^2=-9\to x+4=\pm√(-9)\n\nx+4=-3i\ \vee\ x+4=3i\ \ \ \ |-4\n\nx=-4-3i\ \vee\ x=-4+3i