Answer:
Square shaped = 27 cubic inches , Rectangular shaped = 3LW cubic Inches
Step-by-step explanation:
If the dictionary is a square ( all sides are equal ) the the volume is
V = Length * Length * Length = 
V = 3 * 3 * 3 = 27 
while if its rectangular ( all side are not equal ) shaped
V = Length * Width * Height (thickness)
V = L * W * 3 = 3LW
Answer:
2
Step-by-step explanation:
5-3=2 you subtract to get the answer
Answer:
a. 
b. 
Step-by-step explanation:
The initial value problem is given as:
Applying laplace transformation on the expression 
to get ![L[{y+y'} ]= L[{7 + \delta (t-3)}]](https://tex.z-dn.net/?f=L%5B%7By%2By%27%7D%20%5D%3D%20L%5B%7B7%20%2B%20%5Cdelta%20%28t-3%29%7D%5D)
Taking inverse of Laplace transformation
![y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=y%28t%29%20%3D%207%20L%5E%7B-1%7D%20%5B%20%5Cdfrac%7B1%7D%7B%28s%2B1%29%7D%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207L%5E%7B-1%7D%20%5B%5Cdfrac%7B%28s%2B1%29-s%7D%7Bs%28s%2B1%29%7D%5D%20%2BL%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207L%5E%7B-1%7D%20%5B%5Cdfrac%7B1%7D%7Bs%7D-%5Cdfrac%7B1%7D%7Bs%2B1%7D%5D%20%2B%20L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207%20%5B1-e%5E%7B-t%7D%20%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
![L^{-1}[\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
![L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7B1%7D%7Bs%2B1%7D%5D%20%3D%20e%5E%7B-t%7D%20%20%3D%20f%28t%29%20%5C%20then%20%5C%20by%20%5C%20second%20%5C%20shifting%20%5C%20theorem%3B)
![L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%3D%20%5Cleft%20%5C%7B%20%7B%7Bf%28t-3%29%20%5C%20%5C%20%5C%20t%3E3%7D%20%5Catop%20%7B0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20t%20%3C3%7D%7D%20%5C%20%5C%20%5C%20%20%5Cright.)
![L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%3D%20%5Cleft%20%5C%7B%20%7B%7Be%5E%7B%28-t-3%29%7D%20%5C%20%5C%20%5C%20t%3E3%7D%20%5Catop%20%7B0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20t%20%3C3%7D%7D%20%5C%20%5C%20%5C%20%20%5Cright.)
= 
Recall that:
![y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=y%28t%29%20%3D%207%20%5B1-e%5E%7B-t%7D%20%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
Then
you cant add or subtract or multiply or divide if the variables arent the same so the simplified version of <span>-3x²ym + 7x - 5ymx² + 16x would be
</span> -3x²ym <span>- 5ymx² + 23x</span>
The transformation that shifts a function along the
is
, which shifts the function to the left, and
, which shifts the function to the right.
The transformation that shifts a function along the
is
, which shifts the function up, and
, which shifts the function down
Hence, the correct answer is D:
