Answer:
c. g(x) = 4x^2
Step-by-step explanation:
From a first glance, since g(x), is skinnier than f(x), meaning that it is increasing faster, so I know that I can eliminate options A & B since the coefficient on x needs to be greater than 1.
We can then look and see that g(1) = 4 as shown by the point given to us on the graph.
To find the right answer we can find g(1) for options C & D and whichever one matches the point on the graph is our correct answer. e
Option C:
once we plug in 1 for x, our equation looks like
4(1)^2.
1^2 = 1, and 4(1) = 4,
so g(1) = 4. and our point is (1,4).
This is the same as the graph so this is the CORRECT answer.
If you want to double check, you can still find g(1) for option D and verify that it is the WRONG answer.
Option D:
once we plug in 1 for x, our equation looks like
16(1)^2
1^2 = 1, and 16(1) = 16,
so g(1) = 16. and our point is (1,16).
This is different than the graph so this is the WRONG answer.
A)
Let x represent the cost of 1 student, and y the cost of 1 teacher.
B)
In the first group, there's 25 students and 2 teachers. Their total cost is $97.50
So 25x + 2y = 97.50
In the second group, there's 32 students and 3 teachers. Their total cost is $127
So 32x + 3y = 127
We get the following system of equations:
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)
C)
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)
In equation (1)
25x + 2y = 97.50
25x + 2y - 2y = 97.50 - 2y
25x = 97.50 - 2y
25x / 25 = 97.50/25 - 2y/25
x = 3.9 - (2/25)y
In equation (2), let's replace x by its algebraic value
32x + 3y = 127
32(-2/25y + 3.9) + 3y = 127
11/25y + 124.8 = 127
11/25y + 124.8 - 124.8 = 127 - 124.8
11/25y = 2.2
(11/25y) / (11/25) = 2.2 / (11/25)
y = 5
x = -2/25y + 3.9
x = -2/25 * 5 + 3.9
x = 3.5
So the cost of each student is $3.5, and the cost of each teacher is $5.
Hope this helps! :)
32 is the correct answer I believe if using PEMDAS method
Answer:
200 bars
Step-by-step explanation:
$300 - $50 = the amount of money you still need, $250
250 dollars/$1.25 per bar = 200 bars you need to sell to get 300 dollars
