Answer:
1) as-16a
taknig a as common
=a(s-16)
2)as-4a³
taking a as common
=a(s-4a²)
3)2a³+8a-4a²-16
taking common terms
=2a(a²+4)-4(a²+4)
=(2a-4)(a²+4)
again taking common
=2(a-2)(a²+4)
i hope this will help you :)
Answer:
£1837.5
Step-by-step explanation:
Given data
Cost of car P= £2100.
Rate r= 2.2%
Time t= 6 years
Now we want to find the worth after 6 years, let us apply the compound interest expression but this time for depreciation
A= P(1-r)^t
Substitute
A= 2100(1-0.022)^6
A= 2100*(0.978)^6
A= 2100*0.875
A= £1837.5
Hence the amount of the car after 6 years is £1837.5
Answer:
A
Step-by-step explanation:
v=πr2h
r=(3)²* 5
45π unit³
9514 1404 393
Answer:
64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729
Step-by-step explanation:
The row of Pascal's triangle we need for a 6th power expansion is ...
1, 6, 15, 20, 15, 6, 1
These are the coefficients of the products (a^(n-k))(b^k) in the expansion of (a+b)^n as k ranges from 0 to n.
Your expansion is ...
1(2k)^6(-1/3)^0 +6(2k)^5(-1/3)^1 +15(2k)^4(-1/3)^2 +20(2k)^3(-1/3)^3 +...
15(2k)^2(-1/3)^4 +6(2k)^1(-1/3)^5 +1(2k)^0(-1/3)^6
= 64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729
