Answer:
help me
Step-by-step explanation:
I am having trouble in math please help it's a quiz and I need to pass it
Answer:
Answer: 216 cm2 (square centimetres 
, in your question you had to put cm3, cubic centimetres, it's IMPORTANT )
Step-by-step explanation:
A perfect cube by definition has 3 equal dimensions, as an immediate rule: volume and total surface are equal, only the unit of measure changes (cubic for the volume, square for surface).
But let's calculate it anyway:
Volume = Edge * Edge * Edge = length * width * depth = 
(remember: all edges are equal in this case)
so Edge = ![\sqrt[3]{Volume}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7BVolume%7D)
in your example Edge = ![\sqrt[3]{216}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B216%7D)
= 6cm
So the surface of one side is 6*6 = 36
There are 6 sides in total, so the total surface is 6*36 = 216
Note: I call them "edges" but in case of a cube most say just "length"
Answer:
let me just say this
Step-by-step explanation:
you're luck not mine so sorry if this didnt help
Answer:
if you are being asked what the values of the x and y variables are, x=5/2 and y=4
Step-by-step explanation:
In order to solve this problem you will want to isolate one of the two variables in both equations.
I isolated the y variable in both equations so you have:
4x+y=14, when you isolate the y variable, you will get: y=-4x+14
6x-3y=3, same procedure isolate y: 6x-3=3y, now simplify so you only have y: 2x-1=y.
Now that you have the two equations which a common variable, solve them by making the two equations equal to each other like so:
-4x+14=2x-1, now you want to have all the x variable terms on one side and all the constants on the other, you will add "4x" to each side of the equal sign and then add "1" to each side to leave you with: 15=6x
now divide both sides by 6 to obtain your x value: x=15/6 simplifies to 5/2
then plug your x-value into one of the two equations we made earlier:
y=2x-1
y=2(5/2)-1
y=5-1
y=4
now you have calculated the values of your x and y values
(3v+19)°
the answer to the question
