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.
Answer:
9.65-9.74
Those numbers and anything between them.
(3x) + (x+60) = 180
4x + 60 = 180
Subtract 60 from both sides
4x = 120
Divide 4 from both sides
x = 30
Complete question :
Polly’s Pasta and Pizza Supply sells wholesale goods to local restaurants. They keep track of revenue earned from selling kitchen supplies and food products. The function K models revenue from kitchen supplies and the function F models revenue from food product sales for one year in dollars, where t represents the number of the month (1 – 12) on the last day of the month.
K(t)=15t^3–312t^2+1600t+1100
F(t)=36t3–720t2+3800t–1600
In which month was her revenue from food products the greatest? The least?
Answer:
greatest revenue = 4384
Leat revenue = 400 ; October
Step-by-step explanation:
Using the model :
Claculating the revenue from. Food sales from January through December :
F(t)=36t^3–720t^2+3800t–1600
Put t = 1, 2, 3,.. 12 for January through December
t = 1
F(1)=36(1^3) - 720(1^2) + 3800(1) - 1600 = 1516
t = 2
F(2) = 36(2^3) - 720(2^2) + 3800(2) - 1600 = 3408
Inputting the t values up to 12 to cover for January thkorugh December :
We have the data:
t1 = 1516
t2 = 3408
t3 = 4292
t4 = 4384
t5 = 3900
t6 = 3056
t7 = 2068
t8 = 1152
t9 = 514
t10 = 400
t11 = 996
t12 = 2528
Hence, greatest revenue = 4384 (month of April)
Least revenue is at t10(October) ; Lowest revenue is at t 10 = 400
