Plug in what we know:
Find the cube root of both sides:
![\sf~a=\sqrt[3]{2744}](https://tex.z-dn.net/?f=%5Csf~a%3D%5Csqrt%5B3%5D%7B2744%7D)
Simplify:
Answer:
The expression is
![.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B \ \ \ \ \ \ C \ \ \ \ \ T\\M = \left \ NYC} \atop {LA}} \right. \left[\begin{array}{ccc}{284}&{483}&{410}\\{597}&{551}&{233}{2}}\\\end{array}\right]](https://tex.z-dn.net/?f=.%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20B%20%5C%20%20%20%5C%20%5C%20%5C%20%5C%20%5C%20C%20%5C%20%5C%20%5C%20%5C%20%5C%20T%5C%5CM%20%3D%20%5Cleft%20%5C%20%20NYC%7D%20%5Catop%20%7BLA%7D%7D%20%5Cright.%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B284%7D%26%7B483%7D%26%7B410%7D%5C%5C%7B597%7D%26%7B551%7D%26%7B233%7D%7B2%7D%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
From the question we are told that
The number of each items made in the month of January[J] is
![.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B \ \ \ \ \ C \ \ \ \ \ \ T\\J = \left \ NYC} \atop {LA}} \right. \left[\begin{array}{ccc}{144}&{474 }&{274}\\{598}&{572}&{302}\\\end{array}\right]](https://tex.z-dn.net/?f=.%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20B%20%5C%20%20%5C%20%5C%20%5C%20%5C%20C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20T%5C%5CJ%20%3D%20%5Cleft%20%5C%20%20NYC%7D%20%5Catop%20%7BLA%7D%7D%20%5Cright.%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B144%7D%26%7B474%20%7D%26%7B274%7D%5C%5C%7B598%7D%26%7B572%7D%26%7B302%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
The number of each items made in the month of February[F] is
![.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B \ \ \ \ \ C \ \ \ \ \ \ T\\J = \left \ NYC} \atop {LA}} \right. \left[\begin{array}{ccc}{ 424}&{492}&{546}\\{596 }&{530}&{164}\\\end{array}\right]](https://tex.z-dn.net/?f=.%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20B%20%5C%20%20%5C%20%5C%20%5C%20%5C%20C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20T%5C%5CJ%20%3D%20%5Cleft%20%5C%20%20NYC%7D%20%5Catop%20%7BLA%7D%7D%20%5Cright.%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B%20424%7D%26%7B492%7D%26%7B546%7D%5C%5C%7B596%20%7D%26%7B530%7D%26%7B164%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Generally given that the production for March of all products at all locations was the average of the January and February production then the number of each items made in the month of March[M] is
![.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B \ \ \ \ \ \ \ \ \ \ \ \ \ C \ \ \ \ \ \ \ \ \ \ \ T\\M = \left \ NYC} \atop {LA}} \right. \left[\begin{array}{ccc}{\frac{144 +424}{2}}&{\frac{474 +492}{2}}&{\frac{274 +546}{2}}\\{\frac{598 +596}{2}}&{\frac{572 +530}{2}}&{\frac{302+164}{2}}\\\end{array}\right]](https://tex.z-dn.net/?f=.%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20B%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20T%5C%5CM%20%3D%20%5Cleft%20%5C%20%20NYC%7D%20%5Catop%20%7BLA%7D%7D%20%5Cright.%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B%5Cfrac%7B144%20%2B424%7D%7B2%7D%7D%26%7B%5Cfrac%7B474%20%2B492%7D%7B2%7D%7D%26%7B%5Cfrac%7B274%20%2B546%7D%7B2%7D%7D%5C%5C%7B%5Cfrac%7B598%20%2B596%7D%7B2%7D%7D%26%7B%5Cfrac%7B572%20%2B530%7D%7B2%7D%7D%26%7B%5Cfrac%7B302%2B164%7D%7B2%7D%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=>
Answer:
B. n / 4; 6
Step-by-step explanation:
Let
n = the unknown number
the quotient of a number and four
= n/4
When n = 24
The expression becomes
= n/4
= 24/4
= 6
B. n / 4; 6
2. 3/2x²
-1/2x²
1/4x²
5. -4x²-5 moves in 5 up on the y axis
6. I'm cant remember the trick to this one
7. This one should be D.
Hope This Helps!
"<span>Start by setting up the standard equation for perimeter of a parallelogram, P=2w+2h (these variables of course being width and height). </span>
Now, substitute in what you know...
<span>The problem tells us that width is equal to height minus four. This means that, in order to keep only one variable in this problem, we will write width as h-4. </span>
<span>The problem also tells us that 72 is the perimeter, so we substitute that for P. </span>
72=2(h-4)+2h
<span>Now solve from here. </span>
<span>(Distribute first) </span>
72=2h-8+2h
(Combine like terms)
72=4h-8
80=4h
(Isolate h)
80/4=h
<span>20=h"</span>
