There were 21 students in Travis's fourth-grade class at the end of the year. During the year four new students joined his class, and two moved away. One student was transferred to another fourth-grade teacher. How many students were there in the beginning of the year?

Answers

Answer 1
Answer: so there were 21 students from the start then 4 new people joined so 21 plus 4 equals 25 . 25 minus 2 equals 23 23 minus 1 equals 22
Answer 2
Answer: You start with 21 then add 2 as 2 moved away 21+2=23 1 student was transferred add that on 23+1=24 At the begging of the year there were 24 students.

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There are 5 new books and 7 used books on a shelf. What is the ratio of new books to all books?

Answers

The ratio of new books to all books is 5:12.

In this scenario, there are 5 new books and 7 used books on a shelf. We will determine the ratio of new books to all books and explain it mathematically.

To find the ratio of new books to all books, we need to compare the number of new books to the total number of books, which includes both new and used books. In this case, there are 5 new books and 7 used books, so the total number of books is 5 + 7 = 12.

Now, let's define the ratio of new books to all books using the mathematical notation. We'll use "n" to represent the number of new books, "u" to represent the number of used books, and "t" to represent the total number of books.

Let:

n = 5 (number of new books)

u = 7 (number of used books)

t = n + u = 5 + 7 = 12 (total number of books)

The ratio of new books to all books is given by n/t. Substituting the values, we get:

Ratio of new books to all books = 5 / 12

The ratio can also be expressed in fractional form as 5/12. This means that out of every 12 books on the shelf, 5 are new books, and the remaining 7 are used books. The ratio is simplified to its lowest terms, making it a concise representation of the relationship between new books and all books on the shelf.

To know more about Ratio here

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the answer is 5 over 12.           

Pauling works out with a 2.5-kilogram mass. what is the mass of the 2.5-kilogram mass in grams

Answers

1 kilogram= 1,000 grams 2 kilograms= 2,000 grams 1,000/2=500 2,000+500=2,500 grams 2.5 kilograms= 2,500 grams. Hope this helps!! :)

Sasha has a triangle with vertices A,B and C. The triangle has three congruent angles. She wants to show that triangle ABC has three congruent sides,but she does not have a ruler to measure the lengths of the sides. How can she show that the triangle has three congruent sides?

Answers

She can draw a triangle and then write the lenghts of the triangles with the same numbers

Which expressions are equivalent to 4b ?Choose all answers that apply:
a. b+2 (b+2b)
b. 3b+b
c. 2(2b)

Answers

Answer:

They are b and c.

Step-by-step explanation:

Let's simplify each expression:

a. b+2 (b+2b)

= b + 2b + 4b = 7b.

b.  3b + b = 4b.

c. 2(2b) = 2 * 2b = 4b.

Answer:

hello :

expressions are equivalent to 4b :

b )    3b +b = 4b

c)    2(2b) = 4b

but :  a) is not equivalent to 4b  

because :   b+2 (b+2b) = b + 2b +4b = 7b


What are these fractions in simplest form? ( when I do a / it means a "fraction bar" or " over" like 16/30 or 16 over 30, just incase any of you get confused) but if you could help me simplify these that would be great!!

16y3/20y4 (that's a y to the third power for sixteen and a y to the fourth power for twenty)

6xy/16y

Abc/10abc

Mn2/pm5n (that's an n to the second power for nm and a m to the fifth power on pmn)

12h3k/16h2k2 ( that's a h to the third power for 12hk and a h to the second power and k to the second power for 16hk

8x/10y

24n2/28n ( that's a n to the second power)

30hxy/54kxy

5jh/15jh3 (that's a h to the third power for 15jh)

20s2t3/16st5 ( that's a a to the second power and t to the third power for 20st and t to the fifth power for 16st)

Answers

(16y^(3))/(20y^(4)) =  (4)/(5y)
(6xy)/(16y) =  (3x)/(8)
(abc)/(10abc) =  (1)/(10)
(mn^(2))/(pm^(5)n) =  (n)/(pm^(4))
(12h^(3)k)/(16h^(2)k^(2)) =  (3h)/(4k)
(8x)/(10y) =  (4x)/(5y)
(24n^(2))/(28n) =  (6n)/(7)
(30hxy)/(54kxy)  =  (5h)/(9k)
(5jh)/(15jh^(3)) =  (1)/(3h^(2))
(20s^(2)t^(3))/(16st^(5)) =  (5s)/(4t^(2))

Which angle has its terminal side in the third quadrant?

Answers

Answer:

7π/6.

Step-by-step explanation:

The range in the third quadrant is  π to 3π/2 ( or π to 1.5π) so its not π/3 or 3π/4.

5π/3 = 1.67π so it's not this one.

7π/6 = 1.167π  so it's this one.