first off, let's recall that a cube is just 6 squares stacked up to each other, like in the picture below. Since we know its volume, we can find how long each side is.
part A)
![\bf \textit{volume of a cube}\\\\V=x^3~~\begin{cases}x=side\\[-0.5em]\hrulefill\\V=64\end{cases}\implies 64=x^3\implies \sqrt[3]{64}=x\implies 4=x](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cube%7D%5C%5C%5C%5CV%3Dx%5E3~~%5Cbegin%7Bcases%7Dx%3Dside%5C%5C%5B-0.5em%5D%5Chrulefill%5C%5CV%3D64%5Cend%7Bcases%7D%5Cimplies%2064%3Dx%5E3%5Cimplies%20%5Csqrt%5B3%5D%7B64%7D%3Dx%5Cimplies%204%3Dx%20)
part B)
they have a painting with an area of 12 ft², will the painting fit flat against a side? Well, it can only fit flat if the sides of the painting are the same length or smaller than the sides of the crate, we know the crate is a 4x4x4, so are the painting's sides 4 or less?
![\bf \textit{area of a square}\\\\A=s^2~~\begin{cases}s=side\\[-0.5em]\hrulefill\\A=12\end{cases}\implies 12=s^2\implies \sqrt{12}=s\implies \stackrel{yes}{3.464\approx s}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Barea%20of%20a%20square%7D%5C%5C%5C%5CA%3Ds%5E2~~%5Cbegin%7Bcases%7Ds%3Dside%5C%5C%5B-0.5em%5D%5Chrulefill%5C%5CA%3D12%5Cend%7Bcases%7D%5Cimplies%2012%3Ds%5E2%5Cimplies%20%5Csqrt%7B12%7D%3Ds%5Cimplies%20%5Cstackrel%7Byes%7D%7B3.464%5Capprox%20s%7D%20)
Number of zids=x
number of zods=y
number of zid legs=5x
number of zod legs=7y
so
5x+7y=140
try to get it into slope intercept from so you can graph is (y=mx+b)
5x+7y=140
subtract 5x from both sides
7y=140-5x
divide both sides by 7
y=20-5/7x
y=-5/7x+20
plug in numbers for x and get numbers for y (you can only plug in multiples of 7 for x so that there are a whole number of zids and since you are counting, x and y must never be negative)
lets try 0 for x
y=-5/7(0)+20
y=20
so x=0 and y=20 is an answer (if you can have only one of that species)
lets try 7 for x
y=-5/7(7)+20
y=-5+20
y=15
so x=7 and y=15 is an answer
lets try 14 for x
y=-5/7(14)+20
y=-10+20
y=10
so x=14 and y=10 is another answer
lets try 21 for x
y=-5/7(21)+20
y=-15+20
y=5
so x=21 and y=5 is another answer
lets try 28 for x
y=-5/7(28)-20
y=-20+20
y=0
so x=28 and y=0 is an answer (if there can be only one of a species)
if we go further, then y will be negative so the answers are
(x,y)
(0,20)
(7,15)
(14,10)
(21,5)
(28,0)
if it is allowed that only one species exists then there are 5 possible answers
if both must exist simultaneously, then there are only 3 answers
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Let's solve this problem step-by-step.
STEP-BY-STEP SOLUTION:
We will be using simultaneous equations to solve this problem.
Let's establish the two equations we will be using to solve the problem.
Let Andrew's current age = a
Let Andrew's son's current age = s
Equation No. 1 -
a = 3s
Equation No. 2 -
a - 10 = 5s
To begin with, we will substitute the value of ( a ) from the first equation into the second equation to solve for ( s ).
Equation No. 2 -
a - 10 = 5s
( 3s ) - 10 = 5s
3s - 5s = 10
- 2s = 10
s = 10 / - 2
s = - 5
Next we will substitute the value of ( s ) from the second equation into the first equation to solve for ( a ).
Equation No. 1 -
a = 3s
a = 3 ( - 5 )
a = - 15
FINAL ANSWER:
Therefore, the present age of Andrew is - 15 and the present age of Andrew's son's is - 5.
It isn't possible for someone to be negative years old, but this is the answer that I ibtained from the equations.
Hope this helps! :)
Have a lovely day! <3
Answer:
i took this test along time ago 3,1,1
Answer:
I don't know the answer if you know then tell me i also need the answer
