What is the value of 2x to the forth power + 5x when x =6

(a + b)^n = a^n + n * a^(n - 1) * b + [ n(n - 1)/2! * a^(n - 2) * b^2 ] + + b^n <===let's apply :

(x+3y)^7 = x^7 + 7 * x^6 * 3y + [ (7*6/2) * x^5 * (3y)^2 ] + 7*6*5/3! * x^4 * (3y)^3 + 7*6*5*4/4! * x^3 * (3y)^4 + 7*6*5*4*3/5! * x^2 * (3y)^5 + 7*6*5*4*3*2/6! * x * (3y)^6 + 7*6*5*4*3*2*1/7! * (3y)^7

(x+3y)^7 = x^7 + 21 * x^6 * y + 21 * x^5 * 9y^2 + 7*6*5/6 * x^4 * 27y^3 + 7*6*5*4/24 * x^3 * 81y^4 + 7*6*5*4*3/120 x^2 * 243y^5 + 7*6*5*4*3*2/720 * x * 729y^6 + 1 * 2187y^7

(x+3y)^7 = x^7 + 21x^6 * y + 189 * x^5 * y^2 + 945 * x^4 * y^3 + 2835 * x^3 * y^4 + 5103 x^2 * y^5 + 5103 * x * y^6 + 2187y^7 < that's right, they are 8 terms.

(2x+y)^20 = (2x)^20 + 20 * (2x)^19 * y + ... + y^20 < lat term is y^20

(2x-z)^2)^6

(2x-z)^12 = (2x)^12 + 12 * (2x)^11 * -z + 12*11/2! * (2x)^10 * (-z)^2 + 12*11*10/3! * (2x)^9 * (-z)^3 +

fourth term : 12*11*10/3! * (2x)^9 * (-z)^3 = 220 * 512x^9 * -z^3 = -112640 * x^9 * z^3

i am not sure for this...ain't to the power of 12???

(x+2)^5 = x^5 + 5 * x^4 * 2 + 5 * 4/2! * x^3 * 2^2 + 5 * 4 * 3/3! * x^2 * 2^3 + 5*4*3*2/4! * x * 2^4 + 5*4*3*2*1/5! * x^0 * 2^5

(x+2)^5 = x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 1 * 1 * 32

(x+2)^5 = x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32 <===Answer and there is a missing term in the given options.

(3x + 2y)^5 = (3x)^5 + 5 * (3x)^4 * 2y + 5*4/2! * (3x)^3 * (2y)^2 +

(3x + 2y)^5 = 243x^5 + 810x^4 * y + 1080x^3 * y^2 +

Step-by-step explanation:

x=6 so you have to do 2×6 which equals 12. Then do 5×6 which equals 30. Add that and it equals 42.

Step-by-step explanation:

c. 16 units

step-by-step explanation:

mark brainliest and have a great day!

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this