If the perimeter of the square is 4x then the domain of the function will be set of rational numbers and the domain of the function y=3x+8(3-x) is set of real numbers.
Given The perimeter of the square is f(x)=4x and the function is y=3x+8(3-x)
We will first solve the first part in which we have been given that the perimeter of the square is 4x and we have to find the domain of the function.
First option is set of rational numbers which is right for the function.
Second option is set of whole numbers which is not right as whole number involves 0 also and the side of the square is not equal to 0.
Third option is set of integers which is not right as integers involve negative number also and side of square cannot be negative.
Hence the domain is set of rational numbers.
Now we will solve the second part of the question
f(x)=3x+8(3-x)
we have not told about the range of the function so we can put any value in the function and most appropriate option will be set of real numbers as real number involve positive , negative and decimal values also.
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(8a^5 +2b) + (-6a^5 +2b) = (2a^5 +4b)
hope so much that i ve understood it right theses fives are exponents of ,,a"
Answer
IS 2/4
Step-by-step explanation:
1/3 = 4/12 1/4 = 3/12 4/12+3/12= 6/12 = 2/4
Answer:
-1 5/9
Step-by-step explanation:
so first, simplify the signs so they don’t confuse you,
-5/6 - 17/18 + 2/9 (Since there are two negatives near the end, and their tecxhnically being mulitplied, -2/9*-1 (I added 1 for place holder) and when you multiply a negatives by a negative, you get a positive.
So find the lcm of the denominators which is 18 in this case
-15/18 - 17/18 + 4/18
Now solve the problem from left to right
you’ll get -1 5/9 if you simplified
That would be choice C the limit of f(x) is 2.
