We need to find the expression for " number_of_prizes is divisible number_of_participants". Also there should not remain any remainder left. On in order words, we can say the reaminder we get after division is 0.
Let us assume number of Prizes are = p and
Number of participants = n.
If we divide number of Prizes by number of participants and there will be not remainder then there would be some quotient remaining and that quotent would be a whole number.
Let us assume that quotent is taken by q.
So, we can setup an expression now.
Let us rephrase the statement .
" Number of Prizes ÷ Number of participants = quotient".
p ÷ n = q.
In fraction form we can write
p/n =q ; n ≠ 0.
Answer:
The cost is $9.70 per kilogram.
Step-by-step explanation:
This can be solved by a rule of three.
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too. In this case, the rule of three is a cross multiplication.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease. In this case, the rule of three is a line multiplication.
In this problem, the measures are the weight of the cheese and the price. As the weight increases, so does the price. It means that this is a direct rule of three.
Solution:
The problem states that cheese costs $4.40 per pound. Each kg has 2.2 pounds. How many kg are there in 1 pound. So:
1 pound - xkg
2.2 pound - 1 kg


kg
Since cheese costs $4.40 per pound, and each pound has 0.45kg, cheese costs $4.40 per 0.45kg. How much does is cost for 1kg?
$4.40 - 0.45kg
$x - 1kg



The cost is $9.70 per kilogram.
Answer: it moves at a constant speed and stays in the same direction
Step-by-step explanation:
It does this because there is no friction or nothing pushing on it therefore there is nothing to slow it down or speed it up