Answer:
H. 260
Step-by-step explanation:
We'll begin this problem by first figuring out how many students will be able to sit at the first fourteen tables.
14 tables * 14 students = total students
196 = total students ( for those fourteen tables)
Now we also know that sixteen students can sit on the rest of the cafeteria tables.
We need to find the number of tables can hold sixteen students.
To do this, we'll lead with a simple equation:
18 tables total - 14 tables = # of remaining tables
4 = # of remaining tables
Now we're going to do the same thing we did with the original tables:
4 tables * 16 students = total students
64 = total students
Finally, we add both of the tables max values together:
64 + 196 = 260
Answer : LCM of 2,10,6 is 30 .
You should divide 280 into 3 which would equal 9
Answer:
<u>a) x = 3</u>
<u>b) z = 10</u>
<u>c) p = 2</u>
<u>d) x = 7</u>
<u>e) u = 1</u>
Step-by-step explanation:
a) 2x = 6
Despejamos x dividiendo por 2 a amabos lados de la eacuacion.
(2/2)x = 6/2
<u>x = 3</u>
Si remplazamos x en la ecuación original:
2(3)=6
6 = 6
Queda demostrado.
b) 10 + z = 20
Despejamos z restando 10 en amabos lados de la eacuacion.
10-10+z = 20-10
<u>z = 10</u>
Si remplazamos z en la ecuación original:
10 + 10=20
20 = 20
Queda demostrado.
c) p + 9 = 11
Despejamos p restando 9 en amabos lados de la eacuacion.
p + 9 - 9 = 11-9
<u>p = 2</u>
Si remplazamos p en la ecuación original:
2 + 9 = 11
11 = 11
Queda demostrado.
d) 3x + 8 = 29
Despejamos x restando 8 en amabos lados de la eacuacion y luego divideindo por 3 en ambos lados de la ecuación.
3x+8-8 = 29-8
3x = 21
(3/3)x = 21/3
<u>x = 7</u>
Si remplazamos x en la ecuación original:
3(7) + 8 = 29
21 + 8 = 29
29 = 29
Queda demostrado
e) 2u + 8 = 10
Despejamos u restando 8 en amabos lados de la eacuacion y luego divideindo por 2 en ambos lados de la ecuación.
2u+8-8 = 10-8
2x = 2
(2/2)x = 2/2
<u>x = 1</u>
Si remplazamos x en la ecuación original:
2(1) + 8 = 10
2 + 8 = 10
10 = 10
Queda demostrado
Espero te haya sido de ayuda!