Answer:
let mother's age y and daughter age x
a/q y=3x and y-5=5x-25
put y in second eq we get x=10 &y=30(their present age)
So, the best way to solve this problem is by creating a equation for which we can substitute any number of days. For this problem that equation would be W=5+3(t-1)
W=number of words
T=time or number of days
-1= This is here because we are already accounting for the first day of school so we need to have 1 less day than the number of days they have been in school.
So in 20 days the students had received
5+3(19) or 62words.
Answer: 62 words
Given:
x, y and z are integers.
To prove:
If 
is even, then at least one of x, y or z is even.
Solution:
We know that,
Product of two odd integers is always odd. ...(i)
Difference of two odd integers is always even. ...(ii)
Sum of an even integer and an odd integer is odd. ...(iii)
Let as assume x, y and z all are odd, then is even.
is always odd. [Using (i)]
is always odd. [Using (i)]
is always even. [Using (ii)]
is always odd. [Using (iii)]
is always odd.
So, out assumption is incorrect.
Thus, at least one of x, y or z is even.
Hence proved.
Answer: 2102
Step-by-step explanation:
One metric ton is equal to 1000 Kilograms
