CPhill

This says to "plug" 4 into the function "g" and evaluate the result. Unfortunately, we don't know what "g" is !!!!!

You would have to provide that piece of information before we could answer the question !!!!

Just multiply both sides by 2 sin(.9)...so we have

10*2sin(.9)  = 20sin(.9) = x          and assuming that .9 is in radians, we have.......

x = about 15.66

Most are spot on....!!!!

DragonSlayer554  → Opinionated  (????)

The prime factorization is   2 x 7 x 3²

The formula would be.......

(-1)ⁿ * 8^(n-1)      (where n ≥ 1)

The sum of the interior angles of any triangle measure 180°

So..if all 3 angles are the same, each must measure 60°  .....!!!

Yep.....it factors to 31 x 3.....remember a kind of neat trick....if the sum of the digits of a number is divisible by 3, then that number is also divisible by 3.  So,  93 = 9+3  = 12....and 12 is divisble by 3.... so 93 is, as well!!!

We can also find this by using the difference quotient and the binomial theorem......suppose we want to find the derivative of xⁿ   ...... so we have.....

lim          [ (x + Δx)ⁿ - xⁿ ] / (Δx)   Δx → 0

And by the binomial theorem, we can write

lim          [ xⁿ + C(n, 1)x^n-1 Δx + C(n,2) x^n-2 Δx² + ..... +C(n, n-1) xΔx^n-1 + C(n, n)Δxⁿ  - xⁿ ] / (Δx) 

Δx → 0

lim          [C(n,1) x^n-1 Δx + C(n,2)x^n-2 Δx² + ..... +C(n, n-1) xΔx^n-1 +C(n, n) Δxⁿ ] / (Δx)      

Δx → 0   

.... factor out Δx in the numerator...    

lim         (Δx) [C(n, 1) x^n-1  +C(n, 2) x^n-2 Δx + ..... +C(n, n-1) xΔx^n-2 +C (n, n) Δx^n-1 ] / (Δx)

Δx → 0

lim          [C(n, 1) x^n-1  + C(n, 2) x^n-2 Δx + ..... +C(n, n-1) xΔx^n-2 + C(n, n) Δx^n-1 ]

Δx → 0

And taking the limit as Δx → 0 we have

C(n,1) x ^n-1   ..........  but   C(n, 1) is just n......so we have......

nx^n-1

We have 6 "faces"......with three different rectangular areas being  "doubled"...so we have.......

2 [ (2x * 3x) + (2x * 5x) + (3x * 5x)] = 

2 [ 6x² +  10x² + 15x² ] =

2 [ 31x² ] =

62x²