Answer:
Step-by-step explanation:
All three functions in the figure are parabolic and thus share the parent function . They are horizontally dilated by some constant (unless they aren't) and can all be represented with
, where
is some constant.
Plug in point the line passes through to find that constant. For , we can see the function clearly passes through (2,4):
Therefore, the function of is
Answer:
-2(2x +5)² = -8x² -40x -50
Step-by-step explanation:
Evaluate from the inside out, according to the order of operations.
h(g(f(x))) = h(g(2x +5)) = h((2x +5)²) = -2(2x +5)² = -2(4x² +20x +25)
= -8x² -40x -50
I personally prefer the factored form, but that is not considered "simplified."
Answer: 4) y = -1/3x² + 2x - 1
Step-by-step explanation:
Given :
y = + 2
Expanding (x-3)² will give :
- 6x + 9
substituting into the function , we have :
y = - 6x + 9 ) + 2
Expanding , we have
y = + 2x - 3 + 2
y = + 2x - 1
Answer:
a) There are 11,881,336 of these words in total.
b) There are 7,893,600 of these words with no repeated letters.
c) 896,376 of these words start with an a or end with a z or both
Step-by-step explanation:
Our words have the following format:
L1 - L2 - L3 - L4 - L5
In which L1 is the first letter, L2 the second letter, etc...
There are 26 letters in the English alphabet.
(a) How many of these words are there total?
Each of L1, L2, L3, L4 and L5 have 26 possible options.
So there are of these words total
(b) How many of these words contain no repeated letters?
The first letter can be any of them, so L1 = 26.
At the second letter, the first one cannot be repeated, so L2 = L1 - 1 = 25.
At the third letter, nor the first nor the second one can be repeated, so L3 = L1 - 2 = 24
This logic applies until L5
So we have
26-25-24-23-22
In total there are
of these words with no repeated letters.
(c) How many of these words start with an a or end with a z or both (repeated letters are allowed)?
is the number of words that start with an a and do not end with z. So we have
1 - 26 - 26 - 26 - 25
The first letter can only be a, and the last one cannot be z. So:
is the number of words that start with any letter other than a and end with z. So we have
25 - 26 - 26 - 25 - 1
The first letter can be any of them, other than a, and the last can only be z. So:
is the number of words that both start with a and end with z. So:
1 - 26 - 26 - 26 - 1
The first letter can only be a, and the last can only be z. The other three letters could be anything. So:
896,376 of these words start with an a or end with a z or both