What is the following sum? 2 (RootIndex 3 StartRoot 16 x cubed y EndRoot) 4 (RootIndex 3 StartRoot 54 x Superscript 6 Baseline y Superscript 5 Baseline) 4 x (RootIndex 3 StartRoot 2 y EndRoot) 12 x squared y (RootIndex 3 StartRoot 2 y squared EndRoot) 8 x (RootIndex 3 StartRoot x y EndRoot) 12 x cubed y squared (RootIndex 3 StartRoot 6 y EndRoot) 16 x cubed y (RootIndex 3 StartRoot 2 y squared EndRoot) 48 x cubed y (RootIndex 3 StartRoot 2 y EndRoot).

£56

step-by-step explanation:

u divide 24 by 3 and u get 8 so if u do 8 time 7 or 8+8+8+8+8+8+8 u get 56 and 56 divided by 7 is 8.

it would open into a square depending on the type of envelope i’m not sure about the second part

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.