The constant of proportionality is the ratio between two directly proportional quantities. Two quantities are directly proportional when they increase and decrease at the same rate. The constant of proportionality k is given by k=y/x where y and x are two quantities that are directly proportional to each other.
Solution: The factor form of the given polynomial is 
.
Explanation:
To find the factor form of the given polynomial fisrt find the random value of x for which the vlaue of f(x) is 0.
The value of the function f(x) is 0 for 
, therefore 
is a factor of given function.
Use synthetic method to divide the given polynomial for by . Write coefficients of polynomial in top line and -1 on left side of the line. Write first element in bottom line then multiply it by -1 and write it in second line below the element second element and the add. The division by synthetic method is given in figure 1.
The bottom line shows the coefficients of the quotient polynomial.
So 
Similarly 
is the factor of given polynomial because for x=-2 the value of parenthesis polynomial is 0.
Use synthetic method to divide the parenthesis polynomial for by . it is shown in figure 2.
So
Hence the factor form of the given polynomial is . The graph of the given function is given in figure 3.
Answer:
18 bananas cost the same as 12 apples, and 12 apples cost the same as 8 oranges, so 18 bananas cost the same as $8
I think that should be the answer hope that helped. Let me know if it worked :3
Answer:
-32r-12
Step-by-step explanation:
Solve by distributing the -4 (multiplying -4 by the numbers in parentheses)
Answer: 63lbs
Step-by-step explanation:
The truck can only hold 16crates and each crate weigh 12lbs, the total weight of the 16 crates will be 12×16= 192lbs
This shows that the truck can only contain extra load of 1200lbs - 192lbs = 1008lbs (excluding weight of crates).
To get the shipment weigh close to the total of 1200lbs, the truck must be loaded with engine components not more than 1008lbs.
Since we have 16 crates to fill with engine components not more than 1008lb, each crates will therefore must not exceed 1008/16 pounds of engine components which is equivalent to 63lbs.
Therefore, the manager should instruct the workers to put 63lbs of machine components in "each crate" in order to get the shipment weight as close as possible to 1200 lbs.
