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3 years ago

2p - 1/q = r/q + 4 , subject q

Mathematics

1 answer:

kondaur [170]3 years ago

4 0

Solve for q:

2 p - 1/q = r/q + 4

Bring 2 p - 1/q together using the common denominator q. Bring r/q + 4 together using the common denominator q:

(2 p q - 1)/q = (4 q + r)/q

Multiply both sides by q:

2 p q - 1 = 4 q + r

Subtract 4 q - 1 from both sides:

q (2 p - 4) = r + 1

Divide both sides by 2 p - 4:

Answer: q = (r + 1)/(2 p - 4)

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Read 2 more answers

On the 1st January 2014 Carol invested some money in a bank account.

Ghella [55]

Answer:

$\large \boxed{\text{\pounds 23 360.00}}$

Step-by-step explanation:

The formula for the accrued amount from compound interest is

$A = P \left(1 + \dfrac{r}{n}\right)^{nt}$

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:

r = 0.025

$\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}$

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

$\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}$

3. Summary

1 Jan 2014 P = £23 360.00

1 Jan 2015 A = 23 944.00

Withdrawal = -1 000.00

P = 22 944.00

1 Jan 2016 A = £23 517.60

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3 years ago