Phil ran 2 7/8 miles more than micheal
Answer:
A. yes, the data represents a function because u have no repeating x values. A function cannot have repeating x values...they can have repeating y values, just not the x ones
Step-by-step explanation:
B. table : (8,8)(12,12)(14,16)(16,16)
look at ur points...when x = 8, y = 8...so the table, when x = 8 has a
value of 8
relation : f(x) = 8x - 5....when x = 8
f(8) = 8(8) - 5
f(8) = 64 - 5
f(8) = 59....and the relation has a value of 59
Therefore, the relation has a greater value when x = 8 <==
C. f(x) = 8x - 5...when f(x) = 19
19 = 8x - 5
19 + 5 = 8x
24 = 8x
24/8 = x
3 = x <==
Answer:
2a(b^3 - 7b + 8)
Step-by-step explanation:
I'm assuming that 2a2b3 is 2a2b^2. If not, this answer isn't correct.
Look at the whole numbers. Is there a number that divides into them evenly? Yes, 2, so you pull 2 from the problem and divide each number by 2. Do the same for each variable.
2a2b3 - 14ab + 16a
2(ab^3 - 7ab +8a)
2a(b^3 - 7b + 8)
Answer:
Step-by-step explanation:
To properly apply the substitution method, it will be better for us to rearrange the system of equations to have similar variables on the same side
We can simply evaluate equation 1 to have
y = -20
From the first equation alone, we can evaluate the value of y as -20. This is because only one unknown is present in equation one, hence a single equation is sufficient enough to evaluate it. If to unknowns were present, the two equations would have been utilized to evaluate the solution.
Okay, so 75 counts for 30% (exams), 80 counts for 20% (term paper), and 85 counts for 50% (final exam)
So you would multiply 30% by 75, 20% by 80, and 50% by 85. That would be 22.5, 16, and 42.5. Add these numbers up, and put them over the total percentage, 100%. So you would have 81/100. The student's final average is 81%.
