Let ‘s’ be the son’s age 12 years ago.
Let ‘f’ be the father’s current age.
4 years ago, the son was:
s-4
So, his father is currently:
3(s-4)
=
3s-12
Therefore:
f = 3s-12
In twelve years, the son will be:
s+12
And the father will be:
f+12
This can also be written as:
3s-12+12 as the fathers younger age would be f = 3s+12
=
3s
So, we know that s+12 is half the fathers current age, meaning the father is currently 2(s+12) which is equivalent to 2s+24. Also, we know that the father is currently 3 times the sons age 12 years ago, so 3s (proven by the calculations we made above). Therefore, 2s+24=3s which means 24=s. We can then substitute this, and we will receive 24+12 = 36
Son’s current age: 36
We then substitute the son’s age 12 years ago into 2s+24 to give us the father’s age.
2(24)+24 = 72
Father’s current age: 72
Answer:
A
Step-by-step explanation:
(86°F − 32) × 5/9 = 30°C
They are all prime numbers
Answer:
2 x y^2 (x y - 1) (x y - 3)
Step-by-step explanation:
2x^3y^4-8x^2y^3+6xy^2
each term contains 2xy^2 so factor that out
2xy^2(x^2y^2 -4xy+3)
then lets factor the inside
what terms multiply together to give us +3 and add to -4
-3* -1 = 3 -3+-1 =-4
2 x y^2 (x y - 1) (x y - 3)
Answer:
See Explanation
Step-by-step explanation:
The question is incomplete, as the histogram is not attached.
However, the solution to your question is to select the interval with the highest frequency
Take for instance, after getting readings from the histogram, yiu have:
1 - 10: 8
11 - 20: 5
21 - 30: 11
31 - 40: 2
And so on.......
Interval 11 - 20 will represent the required interval because it has the highest frequency.
