Answer: Choice B. k(h(g(f(x))))
For choice B, the functions are k, h, g, f going from left to right.
===========================================================
Explanation:
We have 4x involved, so we'll need f(x)
This 4x term is inside a cubic, so we'll need g(x) as well.
So far we have
g(x) = x^3
g( f(x) ) = ( f(x) )^3
g( f(x) ) = ( 4x )^3
Then note how we are dividing that result by 2. That's the same as applying the h(x) function
And finally, we subtract 1 from this, but that's the same as using k(x)
This leads to the answer choice B.
To be honest, this notation is a mess considering how many function compositions are going on. It's very easy to get lost. I recommend carefully stepping through the problem and building it up in the way I've done above, or in a similar fashion. The idea is to start from the inside and work your way out. Keep in mind that PEMDAS plays a role.
Rounding to the nearest 10 means if the digit to the right of the 10s collum is 5 or more round up. if less than 5 round down.
605.8
610 is the answer as 5 is rounded up
Answer:
Hey there!
This is because the theoretical probability is different from the experimental probability. The theoretical probability is what is expected, but the experimental probability is what actually happens.
Hope this helps :)
Answer:
welcome❤ though❤ you can be there in the afternoon and you can be there with me
Answer:
Option A, B and E
Step-by-step explanation:
Determinant = ad-bc
Let's look at the picture and solve all
<u><em>Option A)</em></u>
If the row ( c and d ) is zero, the determinant will be zero
=> a(0)-b(0)
=> 0-0
=> 0 (So, True)
<u><em>Option B) </em></u>
If a = b = c = d (Let's say 1), the determinant will be
=> (1)(1)-(1)(1)
=> 1-1
=> 0 (So, True)
<u><em>Option C)</em></u>
An Identity matrix is
=> ![\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
So , it's determinant will be
=> (1)(1)-(0)(0)
=> 1-0
=> 1 (So, False)
<u><em>Option D)</em></u>
The determinant with matrix will all positive numbers can be negative as well as positive. This is not necessary that it would be positive. (So, False)
<u><em>Option E)</em></u>
A zero matrix is
=> ![\left[\begin{array}{ccc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
So, it's determinant is:
=> (0)(0)-(0)(0)
=> 0-0
=> 0 (So,True)
