The sum of all the odd numbers from 1 to 59 is 900.
The method is to multiply
D=dante's cards number
a=anna's cards number
d=60
this (d) is at least (smallest is, aka greater than or equal to or <u>></u>) 6 more than (+6) 3 times as many as anna (3 times a or 3a)
d<u>></u>6+3a
d=60
60<u>></u>6+3a
subtract 6 from both sides
54<u>></u>3a
divide both sides by 3
18<u>></u>a
anna has at least 18 cards
I don't know kid I'm busy right now
Answer:
Step-by-step explanation:
4/3(6y+21)-17=43
move all terms to the left:
4/3(6y+21)-17-(43)=0
Domain of the equation: 3(6y+21)!=0
y∈R
We add all the numbers together, and all the variables
4/3(6y+21)-60=0
You multiply all the terms by the denominator
-60*3(6y+21)+4=0
multiply elements
