Answer:
24 chairs in each class needed
25 classrooms
700 chairs total
how many extra chairs
chairs will be c
t is total
t= 25×c
opposite of multiplication is division( 25 is multiplying by c) c needs to be alone
25 classrooms times chairs equal total chairs
t is 700
700=25×c
700/25=25/25c
700/25=C
c is 28
so 28 chairs for 25 classrooms
they need 24 chairs per class
so there are 4 extra chairs per class
4×25 is 100
**or just multiply 25 classes and 24 chairs
25 times 24 is 600
there are 700 chairs
100 left over
so 100 extra chairs
Assuming 5 is the base. I'm going to leave that out for now.
2log(5x^3) + (1/3)log(x^2+6)
power rule
log(5^2 x^3*2) + log((x^2 + 6)^(1/3))
log(25x^6) + log((x^2 + 6)^(1/3))
quotient rule
log(25x^6 / (x^2 + 6)^(1/3))
Answer:
The answer in the procedure
Step-by-step explanation:
we have
2x-3y=-1 ----> equation A
3x+3y=26 --> equation B
Solve the system by elimination
Adds equation A and equation B
2x-3y=-1
3x+3y=26
----------------
2x+3x=-1+26
5x=25
x=25/5
x=5
Find the value of y
substitute the value of x in equation A or equation B and solve for y
2(5)-3y=-1
10-3y=-1
3y=10+1
y=11/3
I do not understand ur language I speak English very well sorry I can't help you with your homework
