Answer:
-(-45) and -|45|
Step-by-step explanation:
Opposite of a number can also me termed the conjugate of such number. For example, the conjugate (opposite) of a is -a.
Also any value in the modulus sign can either be a negative or positive value. For example |a| can be +a or -a.
According to the question, we are to find the numbers that are equivalent to opposite of -45 i.e which of them are conjugate of -45.
To do this, we are to select the values in the options that will return +45.
Let's look at each option.
For -(-45):
Since the product of two negative sign will give a positive(+), then -(-45) = +45 and hence the value is opposite of -45
For -45:
-45 is a negative value and can never be positive since 45 is not inside a modulus sign. Hence, -45 is not equal to opposite of -45.
For −|−45|, |-45| can only return a positive value since -45 is already in a modulus sign. What the modulus sign will do is convert it to positive value i.e |-45| = 45. Hence -|-45| ≈ -(45) which is not also equivalent to the opposite of-45.
For -|45|:
The value inside modules will return both +45 and -45, hence it can be written as -|45| = -(-45) = +45. This shows that -|45| is also equivalent to the opposite of -45.
Answer:
80
Step-by-step explanation:
Total possible outcomes = 4
Total no. of favorable outcomes = 2

<h2>
Answer:</h2>
(a)
(b)
(c)
<h2>
Step-by-step explanation:</h2>
We are given a function f(x) as :

(a)

We will substitute (x+h) in place of x in the function f(x) as follows:

(b)
Now on subtracting the f(x+h) obtained in part (a) with the function f(x) we have:

(c)
In this part we will divide the numerator expression which is obtained in part (b) by h to get:
Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0