Let A = 0.0009, it is equivalent to A = 9^-4 = (9^-2)^2
so sqrt A =sqrt (9^-2)^2 = (9^-2)
<span>he real square roots of 0.0009 is 0.09</span>
X<20-5y (the whole expression divided by 3)
Answer:
42
Step-by-step explanation:
Given:-
The Sum of the Digit of a Two Digit number is 6.
On reversing, the number is 18 less than the original number.
To Find:-
The original Number.
Let the Ones digit be "x"
ten's digit be = (6 - x)
Original Number = x + 10(6 - x) = 60 - 9x
Now, Atq.
One's Digit = (6 - x)
Ten's digit = x
New Number = 10 × x + (6 - x) = 9x + 6.
So,
(New Number) - (Original Number) = 18
(9x + 6 ) - (60- 9x) = 18
9x + 6 - 60 + 9x = 18
18x - 54 = 18
18x = 18 + 54.
18x = 72
x = 72/18
x = 4.
Therefore,
One's Digit = x = 4
Ten's digit = 6 - x = 2
Original Number = 42
Answer:
The area of the rhombus is 
Step-by-step explanation:
we know that
To find the area of a rhombus, multiply the lengths of the two diagonals and divide by 2
