Answer:
the teacher can make 23 packets with 4 crayons in each packet
Step-by-step explanation:
the teacher can make 23 packets with 4 crayons in each packet wit no left over.
What’s the number then I could help
Answer:
D i believe.
Step-by-step explanation:
P.S. if im wong im really sorry, its been awhile since ive had to do that. But honestly i think its D. also comment on this if im right or wrong pls, id like to know.
Answer:
1) 7m
2) 13
3) 4x^2y
4) 6s^3t^4
Step-by-step explanation:
gcf means the the greatest factor that works for both terms
1) 21m^3 and 28m
let's first find the gcf for the coefficient
21 and 28, the gcf is 7. (hopefully thats easy to explain)
the gcf for the variables is the same mindset as the numbers
m^3 and m, both can be factored by m
so the gcf is 7 * m which is 7m
2)
13x and 26
let's first find the gcf for the coefficient
13 and 26, gcf is 13
there's only one variable, x, so there's not gcf for that
so the gcf is just 13
3)
8x^2y and -12x^3y^2
let's first find the gcf for the coefficient
8 and -12, the gcf is 4 (you can also say -4)
for the variables , same mindset
x^2y and x^3y^2
the gcf for x is x^2, and for y it is just y
so 4 * x^2 * y = 4x^2y
by understanding 3) hopefully you can understand 4)
pls comment for any questions
Step-by-step explanation:
3v/7=6
3v=6×7
3v=42
v=42/3
v=14
hope it helps.
