Suppose the graph of the parent function y=cot(x) is vertically compressed to produce the graph of the function y=a cot(x) but t
2 answers:
Answer:
The required value of a is
Step-by-step explanation:
Given : Suppose the graph of the parent function is vertically compressed to produce the graph of the function
but there are no reflections.
To find : Which describes the value of a?
Solution :
When the graph is vertically compressed by unit a
then the value of a lies between
So, In the graph of parent function is vertically compressed to produce the graph of the function
the value of a lies between
as there is no reflection so no changes.
Therefore, The required value of a is
You might be interested in
Hello! Although I can't give you the answers automatically. I'll just show you some steps on how to subtract & if you're still not figuring it out just message me! :)
Step 1: You have to make sure both denominators are both exactly the same or you won't get the answer correctly.
Step 2: Then subtract the numerators
Step 3: Simplify if needed.
Ex; If I have 1/2 of the pizza and my friend has 1/6 of the pizza. What will be my result if I subtract the fractions?
We'll have to take 1/2 and 1/6. You'll have to find what makes those numbers, what I mean is by multiplying both numerators and denominators.
You multiply by 3.
2 x 3 = 6
Then you'll change the denominator into 1/2 = 1/6
Result:
Then you just subtract.
3 - 1 = 2
Hint: Denominator never changes when you have your final result. So it's 2/6. It needs to be simplified.
We divide by 2.
I hope this example helped you understand how to subtract fractions & if you don't know how to add mixed fractions you can always do around the world or other examples your teacher can provide! :)
Step 1: You have to make sure both denominators are both exactly the same or you won't get the answer correctly.
Step 2: Then subtract the numerators
Step 3: Simplify if needed.
Ex; If I have 1/2 of the pizza and my friend has 1/6 of the pizza. What will be my result if I subtract the fractions?
We'll have to take 1/2 and 1/6. You'll have to find what makes those numbers, what I mean is by multiplying both numerators and denominators.
You multiply by 3.
2 x 3 = 6
Then you'll change the denominator into 1/2 = 1/6
Result:
Then you just subtract.
3 - 1 = 2
Hint: Denominator never changes when you have your final result. So it's 2/6. It needs to be simplified.
We divide by 2.
I hope this example helped you understand how to subtract fractions & if you don't know how to add mixed fractions you can always do around the world or other examples your teacher can provide! :)
DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!
<h2>(a)</h2>Case 1: both balls are white.
At the beginning we have balls. We want to pick a white one, so we have a probability of
of picking a white one.
If this happens, we're left with white balls and still
black balls, for a total of
balls. So, now, the probability of picking a white ball is
The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability
Case 2: both balls are black
The exact same logic leads to a probability of
These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is
Case 1: both balls are white.
In this case, nothing changes between the two picks. So, you have a probability of of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability
of picking a black ball with both picks.
This leads to an overall probability of
Of picking two balls of the same colour.
<h2>(c)</h2>We want to prove that
Expading all squares and products, this translates to
As you can see, this inequality comes in the form
With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:
And this is our case, because in our case we have
so, y has an extra piece and it is larger
which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2