Please answer this math question for me

Mathematics

1 answer:

LiRa [457]3 years ago

4 0

Answer:

Hello! answer: e = 108 d = 32 f = 40 because

108 + 108 + 32 + 32 + 40 + 40 = 360 Hope this helps!

You might be interested in

HELPP!!! 100 POINTS AND BRAINLIEST!! Sketch the graph of the function y = 20x − x2, and approximate the area under the curve in

Arte-miy333 [17]

Answer:

Hello,

Step-by-step explanation:

Part A: see picture

Part B: see picture

Part C:

You must remenber :

$\displaystyle \sum_{i=1}^{n} i= \dfrac{n*(n+1)}{2} \\\\\displaystyle \sum_{i=1}^{n} i^2= \dfrac{n*(n+1)(2n+1)}{6} \\\\\\\int\limits^{20}_0 {(20\ x-x^2)} \, dx =[\dfrac{20\ x^2}{2} -\dfrac{x^3}{3} ]^{20}_0=\dfrac{20^3}{3} =1333.\overline{3}\\\\\\$

We divide the intervalle in n equal parts:

$f(x)=20\ x- x^2\\\Delta x=\dfrac{20-0}{n} =\dfrac{20}{n} \\Area\ of\ a\ rectangle\ R_i=\Delta x*f(x_i)\\x_i=0+\Delta x*i=\Delta x*i\\\\\\\displaystyle \int\limits^{20}_0 {(20 \ x-x^2)} \, dx = \lim_{n \to \infty} \sum_{i=1}^n(\Delta x* f(x_i) )\\\\=\lim_{n \to \infty} \sum_{i=1}^n(\Delta x* (20*\Delta x*i-(\Delta x*i)^2)\\\\=\lim_{n \to \infty} (20*(\dfrac{20}{n})^2\sum_{i=1}^n(i)-(\dfrac{20}{n})^3\sum_{i=1}^n(i^2)\\\\$

$\displaystyle =\lim_{n \to \infty}(20*(\dfrac{20}{n})^2*\dfrac{n*(n+1)}{2} -(\dfrac{20}{n})^3*\dfrac{n*(n+1)(2n+1)}{6} \\\\=10*20^2-\dfrac{20^3}{6} *2\\\\\\=4000-\frac{8000}{3} \\\\=\dfrac{4000}{3} \\\\=1333,\overline{3}\\\\$