The minimal completion time for the activities is the shortest possible time for all the activities to be finished. In doing this, we look at the path that would require the greatest amount of time. At the START node, we choose the path that would take the longest which is 7 days leading to ACTIVITY D. Next, we choose the path leading to ACTIVITY B which takes 5 days. Then, we move to ACTIVITY C taking 5 days and finally, reach the END which would take 6 days. So, the minimal completion time is:
7 + 5 + 5 + 6 = 23 days

6 0

3 years ago

Can someone do eight nine one and two ?

Sergeeva-Olga [200]

Answer: hello there here are your answers:

8) B multiplication property of zero

9) C additive identify

1) 8

2) -2a-7

Step-by-step explanation:

1) $\frac{3+u}8^{2} \\u=5 \\so\\ \frac{3+5}8^{2} \\3+5=8^{2} \\8^{2} =\\64 \\\\ 64/8=\\\\\\8 \\there.\\\\\\$

2) $-2(a-7)\\\\-2(a)(-7)\\\\=-2a+14\\\\\\there$

5 0

3 years ago

Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1

belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

$F(x) = \int\limits^a_b {f(x)} \, dx$

By using integration formula

$\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c$

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

$\int\limits^6_^-6} (x^{5} + 8 x^{4} + 2 x^{2} + 5 x + 15) )dx$

On integration , we get

= $(\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}$

$F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)$

= $(\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))$