<h2>Answer </h2>
Amount (A) = P[1 + (r/100)]n
Principal (P) = ₹ 26400
Time period (n) = 2 years 4 months
Rate % (R) = 15% compounded annually
<h3>Steps </h3>
First, we will calculate Compound Interest (C.I) for the period of 2 years
A = P[1 + (r/100)]n
= 26400[1 + (15/100)]²
= 26400[(100/100) + (15/100)]²
= 26400 × 115/100 × 115/100
= 26400 × 23/20 × 23/20
= 26400 × 1.3225
= 34914
C.I. = A - P
= 34914 - 26400
= 8514
Now, we will find Simple Interest (S.I) for the period of 4 months
Principal for 4 months after C.I. for 2 years = ₹ 34,914
<h3>We know that ,</h3>
S.I = PRT/100
Here T = 4 months = 4/12 years = 1/3 years
S.I. for 4 months = (1/3) × 34914 × (15/100)
= (1/3) × 34914 × (3/20)
= 34914/20
= 1745.70
Total interest for 2 years 4 months = 8514 + 1745.70
= 10259.70
Total amount for 2 years 4 months = 26400 + 10259.70
= ₹ 36659.70
<h3>
So , the correct answer is ₹ 36659.70 . </h3>
Answer:
The teacher disliked her average student because he was like all the other students???
We take the equation <span>d = -16t^2+12t</span> and subtract d from both sides to get
0<span> = -16t^2+12t - d
We apply the quadratic formula to solve for t. With a = -16, b = 12, c = -d, we have
t = [ -(12) </span><span>± √( 12^2 - 4(-16)(-d) ) ] / [2 * -16]</span>
= [- 12 ± √(144-64d) ] / (-32)
= [- 12 ± √16(9-4d)] / (-32)
= [- 12 ± 4√(9-4d)] / (-32)
= 3/8 ±√(9-4d) / 8
The answer to your question is t = 3/8 ±√(9-4d) / 8
Answer:
x=34
Step-by-step explanation:
6 - ( x-7) ^ 1/3 = 3
Subtract 6 from each side
6-6 - ( x-7) ^ 1/3 = 3-6
- ( x-7) ^ 1/3 = -3
Divide each side by a negative
( x-7) ^ 1/3 = 3
Cube each side
( x-7) ^ 1/3 ^3 = (3)^3
x-7 = 27
Add 7 to each side
x-7+7 = 27+7
x = 34
Check
6 - ( 34-7) ^ 1/3 = 3
6 - (27^1/3 = 3
6 -3 =3
3=3
Good solution
Its Your option B.
i.e. 1.325 * 10 ^ 6
