Answer:
$28
Step-by-step explanation:
cp=20
mp=20%ofcp
=20/100*20
=8
now,
sp=mp+cp
=8+20
=28
Answer:
![(AB)^T = B^T.A^T (Proved)](https://tex.z-dn.net/?f=%28AB%29%5ET%20%3D%20B%5ET.A%5ET%20%20%28Proved%29)
Step-by-step explanation:
Given (AB )^T= B^T. A^T;
To prove this expression, we need to apply multiplication law, power law and division law of indices respectively, as shown below.
![(AB)^T = B^T.A^T\\\\Start, from \ Right \ hand \ side\\\\B^T.A^T = \frac{B^T.A^T}{A^T}.\frac{B^T.A^T}{B^T} (multiply \ through) \\\\ = \frac{A^{2T}.B^{2T}}{A^T.B^T} \\\\=\frac{(AB)^{2T}}{(AB)^T} \ \ (factor \ out \ the power)\\\\= (AB)^{2T-T} \ (apply \ division \ law \ of \ indices; \ \frac{x^a}{x^b} = x^{a-b})\\\\= (AB)^T \ (Proved)](https://tex.z-dn.net/?f=%28AB%29%5ET%20%3D%20B%5ET.A%5ET%5C%5C%5C%5CStart%2C%20from%20%5C%20Right%20%5C%20hand%20%5C%20side%5C%5C%5C%5CB%5ET.A%5ET%20%3D%20%5Cfrac%7BB%5ET.A%5ET%7D%7BA%5ET%7D.%5Cfrac%7BB%5ET.A%5ET%7D%7BB%5ET%7D%20%28multiply%20%5C%20through%29%20%5C%5C%5C%5C%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%3D%20%5Cfrac%7BA%5E%7B2T%7D.B%5E%7B2T%7D%7D%7BA%5ET.B%5ET%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B%28AB%29%5E%7B2T%7D%7D%7B%28AB%29%5ET%7D%20%5C%20%5C%20%28factor%20%5C%20out%20%5C%20the%20power%29%5C%5C%5C%5C%3D%20%28AB%29%5E%7B2T-T%7D%20%20%5C%20%28apply%20%5C%20division%20%5C%20law%20%5C%20of%20%5C%20indices%3B%20%5C%20%5Cfrac%7Bx%5Ea%7D%7Bx%5Eb%7D%20%3D%20x%5E%7Ba-b%7D%29%5C%5C%5C%5C%3D%20%28AB%29%5ET%20%5C%20%28Proved%29)
Answer:
-319
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
Here, we want to find which of the expressions have the greatest rate of exponential growth.
The easiest way to go about this is have a substitution for the term t;
Let’s say t = 6
Thus;
h(t) = 1.18^1 = 1.18
K(t) = 0.375^6 = 0.002780914307
f(t) = 1.36^6 = 6.327518887936
g(t) = 0.86^6 = 0.404567235136
Another way to find this is to express each as a sum of 1
f(t) = (1+ 0.36)^t
g(t) = (1-0.14)^t
k(t) = (1-0.625)^t
We can see clearly that out of all the terms in the brackets asides 1, 0.36 is the biggest in value