V = B · h; where B is the area of the base
150 in³ = 30 in² · h
5 in = h
Answer: 5 in
Answer:
z(max) = 256000 Php
x₁ = 10
x₂ = 110
Step-by-step explanation:
Jogging pants design Selling Price Cost
weekly production
Design A x₁ 2500 1750
Design B x₂ 2100 1200
1. z ( function is : )
z = 2500*x₁ + 2100*x₂ to maximize
First constraint weekly production
x₁ + x₂ ≤ 120
Second constraint Budget
1750*x₁ + 1200*x₂ ≤ 150000
Then the model is
z = 2500*x₁ + 2100*x₂ to maximize
Subject to
x₁ + x₂ ≤ 120
1750*x₁ + 1200*x₂ ≤ 150000
General constraints x₁ ≥ 0 x₂ ≥ 0 both integers
First table
z x₁ x₂ s₁ s₂ cte
1 -2500 -2100 0 0 0
0 1 1 1 0 = 120
0 1750 1200 0 1 = 150000
Using AtoZmath online solver and after 6 iterations the solution is:
z(max) = 256000 Php
x₁ = 10
x₂ = 110
Answer:
Tap A 3hrs
Tap B 6hrs
Step-by-step explanation:
Let the volume of the swimming pool be Xm^3.
Now, to get the appropriate volume, we know we need to multiply the rate by the time. Let the rate of the taps be R1 and R2 respectively, while the time taken to fill the swimming pool be Ta and Tb respectively.
x/Ta= Ra
x/Tb= Rb
X/(Ra + Rb)= 2
Ta = Tb - 3
From equation 2:
X = 2( Ra + Rb)
Substituting the values of Ra and Rb Using the first set of equations
X = 2( x/Ta + x/Tb)
But Ta = Tb - 3
1/2 = 1/(Tb - 3)+ 1/Tb
0.5 = (Tb + Tb-3)/Tb(Tb - 3)
At this juncture let’s say Tb = y
0.5 = (2y - 3)/y(y - 3)
y(y-3 ) = 4y - 6
y^2 -3y - 4y + 6 = 0
y^2 -7y + 6= 0
Solving the quadratic equation, we get y =
y = Tb = 6hrs or 1hr
We remove one hour as we know that Tap A takes 3hrs left than tap B and there is nothing like negative hours
Now, we get Ta by Tb -3 = 6 - 3 = 3hrs
Answer:
132 students
Step-by-step explanation:
Mr Smith offers to take extra help session for students after school
The day after the help session 11 of his students were talking and found out that only 2 had stayed up for the help session
= 2/11
There were actually a total number of 24 students that stayed
We are asked to find the number of student that Mr Smith teaches throughout the day
Let x represent the number of students that were taught during the day
Therefore, the number of student that Mr Smith teaches during the day can be calculated as follows
2/11=24/x
Cross multiply both sides
2 × x = 24 × 11
2x= 264
Divide both sides by 2
2x/2=264/2
x= 132
Hence Mr Smith teaches 132 students during the day.
Answer:
2/5
Step-by-step explanation:
There is an 8/20 chance that the next person will receive a window seat. You simplify that probability to get your answer. 8/20 becomes 2/5. So the answer is 2/5.
