Expressions equivalent to 3(1+x)+7 is x+10
Well you need first 1 2/4 of a yard. Then you will need 5/6 of a yard. But first you need to put them together. So we need to find the common denominator. Six and four both share which multiple? Twelve. So let's rewrite that. You need 1 6/12 yards + 10/12 yards of wrapping paper. But before we do anything else we need to fix the mixed number. 1 6/12 isn't as easy to add ad 18/12 is. So 18/12 + 10/12 = 28/12
That can simplify as well. Now it's 14/6 which is equivalent to the fraction we had before. But it can also go down to 2 2/6. Now we can simplify it even further. By dividing it by two we can get 2 1/3. These are the measurements we need. So you would need how many thirds? Seven. (2 1/3 is seven thirds) This is your final answer.
You would need to have seven peices of wrapping paper cut to wrap both gift boxes.
Answer:
A triangle has side lengths 18, 24, x.
18 + 24 = x
18 squared + 24 = x
(18 + 24) squared = x squared
18 squared + 24 squared = x squared
Step-by-step explanation:
Answer:
1/16
Step-by-step explanation:
In the starting, Corrine had 1/4 gallon of blackberries. Now, she poured the berries equally into 4 containers. Therefore, the amount of berries in each container is:1/4 : 4=1/16
Thus, each container has 1/16 of a gallon of blackberries.
Answer:
or 
(simplified)
Step-by-step explanation:
Based on the information provided within the question it can be said that in order to calculate the probability of both grapes being green we need to find the probability of each grape being green separately and then multiply those probabilities together
In the first choice, there are a total of 22 grapes (9+13), 9 of which are green. Therefore the probability of the first chosen grape being green is 
In the second choice,since we removed one grape there is now a total of 21 grapes (22-1), 8 of which are green. Therefore the probability of the second chosen grape being green is 
Now we multiply both probabilities together to calculate the probability that both grapes are green in a sequence.
