Answer:
y = 3
Step-by-step explanation:
y = (3x² + 3x + 6) / (x² + 1)
The power of the numerator and denominator are equal, so as x approaches infinity, y approaches the ratio of the leading coefficients.
y = 3/1
What is the maximum number of chairs and tables Ben can make in a month.
Answer:
Maximum number of tables he can make per month = 100 tables
Maximum number of chairs he can make per month = 400 chairs
Step-by-step explanation:
He is not able to make more than 500 pieces of furniture per month
Also, he needs to make at least 25 tables and at least 100 chairs each month.
x is number of tables and y is number of chairs per month.
Thus, inequality to represent this is;
x + y ≤ 500
Constraints are;
x ≥ 25
y ≥ 100
Since minimum is 25 tables and 100 chairs, we say that 1 table would need 100/25 = 4 chairs
Thus, ratio of tables to chairs should always be 1:4
Thus,
Maximum number of tables per month = 1/5 × 500 = 100 tables
Maximum number of chairs per month = 4/5 × 500 = 400 chairs
If R = 5z
Then 15 z = 3R so there given 15z=3Y soo R= 5 Y
E=number of eagles
c=number of condors
ttoal is 480
c+e=480
5 times as many condors and 9 times as many eagles totals 3260
5c+9e=3260
we have
c+e=480
5c+9e=3260
eliminateion
elmiminate c
times first equation by -5
-5c-5e=-2400
add to other equation
-5c-5e=-2400
<u> 5c+9e=3260 +</u>
0c+4e=860
4e=860
divide both sides by 4
e=215
sub back
c+e=480
c+215=480
minus 215 from both sides
c=265
thiis spring is
265 condors
215 eagles
Answer:
- 30 sets of model B
- 20 sets of model A
Step-by-step explanation:
To maximize profit, the greatest possible number of the most profitable item should be manufactured. Remaining capacity should be used for the less-profitable item.
Up to 30 of model B, which has the highest profit, can be made each day. The remaining amount (20 sets) of the daily capacity of 50 sets should be used to make model A sets.