Answer:
h(8q²-2q) = 56q² -10q
k(2q²+3q) = 16q² +31q
Step-by-step explanation:
1. Replace x in the function definition with the function's argument, then simplify.
h(x) = 7x +4q
h(8q² -2q) = 7(8q² -2q) +4q = 56q² -14q +4q = 56q² -10q
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2. Same as the first problem.
k(x) = 8x +7q
k(2q² +3q) = 8(2q² +3q) +7q = 16q² +24q +7q = 16q² +31q
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Comment on the problem
In each case, the function definition says the function is not a function of q; it is only a function of x. It is h(x), not h(x, q). Thus the "q" in the function definition should be considered to be a literal not to be affected by any value x may have. It could be considered another way to write z, for example. In that case, the function would evaluate to ...
h(8q² -2q) = 56q² -14q +4z
and replacing q with some value (say, 2) would give 196+4z, a value that still has z as a separate entity.
In short, I believe the offered answers are misleading with respect to how you would treat function definitions in the real world.
The answer is
27/5 = 5 2/5 = 5.4
Answer:
5 2/10
Step-by-step explanation:
A=24
Find the area by multiplying both diagonals and dividing them by 2. Like so: And you might be wondering how to do that. By using the pythagorean theorem and some simple addition, you could get the answer. One diagonal is 8 and the other is 6....so, 8x6=48....48/2=24
Please let me know if you have any other questions!
Answer:
- The two solutions are:
- The next and every step are below.
Explanation:
1. 
: Given (addition property / add - 3 to both sides)
2. 
: Given (commom factor - 2)
3. 
To obtain the perfect square it was added the square of half of the coefficient of x: (1/2)² = 1/4, inside the parenthesis.
Since, the terms inside the parentthesis are multiplied by - 2, you have to add - 2 (1/4) = - 1/2 to the left side of the equation.
4. Now, you have that the trinomial x² - x + 1/4 is a square perfect trinomial which is factored as (x - 1/2)² and get the expression:
5. Divide both sides by - 2 to get the next expression:
6. The last step is to extract squere root from both sides of the equality:
