Let, the numbers = x,y
Now, x + y = 38
x - y = 50, x = 50 + y
Substitute it into first equation,
50 + y + y = 38
2y = 38 - 50
y = -12 / 2
y = -6
Substitute it into first equation,
x - 6 = 38
x = 38 + 6 = 44
So, your final answer is x = 44 & y = -6
Hope this helps!
Step-by-step explanation:
Both of them had $x
Later
Jensen=x-64
Raju=x-148
Jensen=3*Raju
x-64=3(x-148)
x-64=3x-444
3x-x=444-64
2x=380
x=$190
∆ABD is right angled hence area:-
There is only one option containing 6x^2 i.e Option D.
Hence without calculating further
Option D is correct
So (7.4*105)-(8.9*103) (The sum of all departments)-(The sum of the accounting department)
(7.4*105)-(8.9*103) = (777)-(916.7) =(-139.7), not sure, did you write the problem wrong? If the total of the company is less than what the account dept used, something is off.
Answer: 10, 11, & 12
<u>Step-by-step explanation:</u>
Let x represent the age of the youngest child.
Their ages are consecutive so,
Youngest: x
Middle: x + 1
Oldest: x + 2
The age of the Youngest squared (x²) equals 8 times the Oldest [8(x + 2)] plus 4.
x² = 8(x + 2) + 4
x² = 8x + 16 + 4
x² = 8x + 20
x² - 8x - 20 = 0
(x - 10)(x + 2) = 0
x - 10 = 0 or x + 2 = 0
x = 10 or x = -2
Since age cannot be negative, x = -2 is not valid
So, the Youngest (x) is 10
the Middle (x + 1) is 11
and the Oldest (x + 2) is 12
