At Mr. McBerry’s farm, new fruit trees are being planted. There are 160 pear trees, 224 peach trees, and 320 plum trees. Find the maximum number of trees that can be planted in each row if Mr. McBerry wants the same number of trees in each row.(1 point)
I apologize for the mistake. Let's find the correct answer.
To find the maximum number of trees that can be planted in each row, we need to find the greatest common divisor (GCD) of the given numbers.
First, let's write the given numbers as a product of their prime factors:
160 = 2^5 * 5
224 = 2^5 * 7
320 = 2^6 * 5
Now, let's identify the common prime factors and their lowest exponents:
Both 160 and 224 have common factors of 2^5 and 1 factor of 7.
Both 160 and 320 have common factors of 2^5 and 1 factor of 5.
Therefore, the greatest common divisor (GCD) of 160, 224, and 320 is 2^5 * 1 * 1 = 32.
This means that the maximum number of trees that can be planted in each row is 32.
wrong
right
Thank you for your understanding. If you have any more questions, feel free to ask.
Find the GCF of 78 and 86.(1 point)
Responses
1
1
2
2
86
86
3
The GCF (Greatest Common Factor) of 78 and 86 is 2.
A rabbit is packing lunches for his children. He has 90 carrots and 72 leaves of lettuce. If he is giving each bunny the same number of carrots and the same number of leaves of lettuce, how many baby bunnies must he have? He will have no extra vegetables after he packs the lunches.(1 point)
Responses
9
9
3
3
6
6
18
To find the number of baby bunnies, we need to find the greatest common factor (GCF) of 90 and 72.
First, let's find the prime factorization of each number:
90 = 2 * 3^2 * 5
72 = 2^3 * 3^2
Next, we identify the common prime factors and their lowest exponents:
2 * 3^2
Finally, we multiply the common prime factors together: 2 * 3^2 = 18.
Therefore, the rabbit must have 18 baby bunnies in order to give each bunny the same number of carrots and the same number of leaves of lettuce.
A carnival game gives variety bags as prizes. The game operator uses 100 balloons and 68 stickers to put into a certain number of bags. How many balloons and how many stickers can go in each bag if they make the greatest number of bags possible so that each bag has the same number of balloons and the same number of stickers?(1 point)
Responses
17 balloons and 25 stickers
17 balloons and 25 stickers
50 balloons and 34 stickers
50 balloons and 34 stickers
25 balloons and 17 stickers
25 balloons and 17 stickers
4 balloons and 4 stickers
To find the maximum number of bags and how many balloons and stickers can go in each bag, we need to find the greatest common divisor (GCD) of 100 and 68.
Let's find the prime factorization of each number:
100 = 2^2 * 5^2
68 = 2^2 * 17
Now, let's identify the common prime factors and their lowest exponents:
2^2
Therefore, the GCD of 100 and 68 is 2^2, which is 4.
This means that each bag would contain 4 balloons and 4 stickers.
Therefore, the correct response is, "4 balloons and 4 stickers."