Answer:
The function y = sec(x) shifted 3 units left and 7 units down .
Step-by-step explanation:
Given the function: y = sec(x)
- If k is any positive real number, then the graph of f(x) - k is the graph of y = f(x) shifted downward k units.
- If p is a positive real number, then the graph of f(x+p) is the graph of y=f(x) shifted to the left p units.
The function
comes from the base function y= sec(x).
Since 3 is added added on the inside, this is a horizontal shift Left 3 unit, and since 7 is subtracted on the outside, this is a vertical shift down 7 units.
Therefore, the transformation on the given function is shifted 3 units left and 7 units down
Answer:
yes because every time x changes y changes the same amount
Step-by-step explanation:
Answer:
formula: a squared + b squared= c squared
6 squared+ 8 squared= c squared
36+64= c squared
100= c squared
Square root both sides
You get 10= c
Given Equations
- 5x+3y=8--(1)
- 4x+7y=34--(2)
Let it has solution (x,y)
Add both


It will also have same solution (x,y)
Answer:
The required probability is 0.94
Step-by-step explanation:
Consider the provided information.
There are 400 refrigerators, of which 40 have defective compressors.
Therefore N = 400 and X = 40
The probability of defective compressors is:

It is given that If X is the number among 15 randomly selected refrigerators that have defective compressors,
That means n=15
Apply the probability density function.

We need to find P(X ≤ 3)


Hence, the required probability is 0.94