2/3 x 3/4 =6/12=1/2
(Multiply numerators, multiply denominators, simplify)
So distribute using distributive property
a(b+c)=ab+ac so
split it up
(5x^2+4x-4)(4x^3-2x+6)=(5x^2)(4x^3-2x+6)+(4x)(4x^3-2x+6)+(-4)(4x^3-2x+6)=[(5x^2)(4x^3)+(5x^2)(-2x)+(5x^2)(6)]+[(4x)(4x^3)+(4x)(-2x)+(4x)(6)]+[(-4)(4x^3)+(-4)(-2x)+(-4)(6)]=(20x^5)+(-10x^3)+(30x^2)+(16x^4)+(-8x^2)+(24x)+(-16x^3)+(8x)+(-24)
group like terms
[20x^5]+[16x^4]+[-10x^3-16x^3]+[30x^2-8x^2]+[24x+8x]+[-24]=20x^5+16x^4-26x^3+22x^2+32x-24
the asnwer is 20x^5+16x^4-26x^3+22x^2+32x-24
Answer:

Step-by-step explanation:
The formula for the accrued amount from compound interest is

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

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.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

3. Summary
1 Jan 2014 P = £23 360.00
1 Jan 2015 A = 23 944.00
Withdrawal = <u> -1 000.00
</u>
P = 22 944.00
1 Jan 2016 A = £23 517.60
The answer is 1.2 hope this helped ;)