Anyone know collage level algebra 2? I need help Evaluate the series: image Question 1 options: A) 392 B) 249 C) 420 D) 406

The required sum is 392

Step-by-step explanation:

Given the sum as shown;

When n = 1

f(1) = 3(1)²-4

f(1) = 3 - 4

f(1) = - 1

When n = 2

f(2) = 3(2)²-4

f(2) = 12 - 4

f(2) = 8

When n = 3

f(3) = 3(3)²-4

f(3) = 27 - 4

f(3) = 23

When n = 4

f(4) = 3(4)²-4

f(4) = 48 - 4

f(4) = 44

When n = 5

f(5) = 3(5)²-4

f(5) = 75 - 4

f(5) = 71

When n = 6

f(6) = 3(6)²-4

f(6) = 108 - 4

f(6) = 104

When n = 7

f(7) = 3(7)²-4

f(7) = 147 - 4

f(7) = 143

Taking their sum

Sum = f(1)+f(2)+(3)+f(4)+f(5)+f(6)+f(7)

Sum = -1 + 8 + 23 + 44 + 71 + 104 + 143

Sum = 392

Hence the required sum is 392

phosphate, calcium ion, chloride ion, neon atom

step-by-step explanation:

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2