Answer: A. 40
Step-by-step explanation:
Answer:
Step-by-step explanation:e equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-factor property is then used to find solutions. ... Another method for solving quadratics is the square root property. The variable is squared.
Answer:
£510
Step-by-step explanation:
Simple interest = (principal × rate × time) / 100
Principal = £1700
Rate = 10%
Time = 3 years
Simple interest = (principal × rate × time) / 100
= (1700 × 10 × 3) / 100
= 51,000/100
= 510
Simple interest = £510
The problem can be solved step by step, if we know certain basic rules of summation. Following rules assume summation limits are identical.
![\sum{a+b}=\sum{a}+\sum{b}](https://tex.z-dn.net/?f=%5Csum%7Ba%2Bb%7D%3D%5Csum%7Ba%7D%2B%5Csum%7Bb%7D)
![\sum{kx}=k\sum{x}](https://tex.z-dn.net/?f=%5Csum%7Bkx%7D%3Dk%5Csum%7Bx%7D)
![\sum_{r=1}^n{1}=n](https://tex.z-dn.net/?f=%5Csum_%7Br%3D1%7D%5En%7B1%7D%3Dn)
![\sum_{r=1}^n{r}=n(n+1)/2](https://tex.z-dn.net/?f=%5Csum_%7Br%3D1%7D%5En%7Br%7D%3Dn%28n%2B1%29%2F2)
Armed with the above rules, we can split up the summation into simple terms:
![\sum_{r=1}^n{40r-21n+8}=n](https://tex.z-dn.net/?f=%5Csum_%7Br%3D1%7D%5En%7B40r-21n%2B8%7D%3Dn)
![=40\sum_{r=1}^n{r}-21n\sum_{r=1}^n{1}+8\sum_{r=1}^n{1}](https://tex.z-dn.net/?f=%3D40%5Csum_%7Br%3D1%7D%5En%7Br%7D-21n%5Csum_%7Br%3D1%7D%5En%7B1%7D%2B8%5Csum_%7Br%3D1%7D%5En%7B1%7D)
![=40\frac{n(n+1)}{2}-21n^2+8n](https://tex.z-dn.net/?f=%3D40%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D-21n%5E2%2B8n)
![=20n(n+1)-21n^2+8n](https://tex.z-dn.net/?f=%3D20n%28n%2B1%29-21n%5E2%2B8n)
![=28n-n^2](https://tex.z-dn.net/?f=%3D28n-n%5E2)
=> (a)
f(x)=28n-n^2=> f'(x)=28-2n
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0 therefore f(14) is a maximum
(b)
f(x) is a maximum when n=14
(c)
the maximum value of f(x) is f(14)=196