Atul has 2/3 lb of candy. Jose has 3/5 lb and Maria has 1/2 lb less then José. How many more pounds of candy does Atul have than
1 answer:
Given:
Atul has lb of candy.
Jose has lb of candy.
Maria has lb less than Jose.
To find:
How many more pounds of candy does Atul have than Maria?
Solution:
Since, Maria has lb less than Jose, therefore
Maria has = lb
Maria has = lb
Maria has = lb
Difference between the candies Atul and Maria have.
Difference = lb
Difference = lb
Difference = lb
Therefore, Atul have lb of candy more than Maria.
You might be interested in
For this case, the parent function is given by:
We apply the following function transformation:
Vertical compressions:
To graph y = a * f (x)
If 0 <a <1, the graph of y = f (x) is compressed vertically by a factor a. (Shrinks)
We have then:
Horizontal translations:
Suppose that h> 0
To graph y = f (x + h), move the graph of h units to the left.
We have then:
Vertical translations:
Suppose that k> 0
To graph y = f (x) + k, move the graph of k units up.
We have then:
Answer:
Vertical compression by factor of 1/2. Horizontal displacement 3 units to the left. Vertical displacement 2 units up.
We apply the following function transformation:
Vertical compressions:
To graph y = a * f (x)
If 0 <a <1, the graph of y = f (x) is compressed vertically by a factor a. (Shrinks)
We have then:
Horizontal translations:
Suppose that h> 0
To graph y = f (x + h), move the graph of h units to the left.
We have then:
Vertical translations:
Suppose that k> 0
To graph y = f (x) + k, move the graph of k units up.
We have then:
Answer:
Vertical compression by factor of 1/2. Horizontal displacement 3 units to the left. Vertical displacement 2 units up.