You can set up an inequality to solve this. You can either do big to small --> 
or small to big --> 
. I'm going to be using big to small because I think it's a little easier.
Cross multiply and you will get:
Isolate x:
x = 
x = 45
The goal post is 45 feet tall. Hope this helped. Good luck. : )
R(x) = 60x - 0.2x^2
The revenue is maximum when the derivative of R(x) = 0.
dR(x)/dx = 60 - 0.4x = 0
0.4x = 60
x = 60/0.4 = 150
Therefore, maximum revenue is 60(150) - 0.2(150)^2 = 9000 - 4500 = $4,500
Maximum revenue is $4,500 and the number of units is 150 units
Answer:Let the two unknown numbers be x and y.
So, x-y=40 ........Equation 1
And 0.3(x)= 37.5/100 (y)
From equation 1, x=40+y
Now, multiply through by 100 in equation 2.
We have,
30x = 37.5 (y)
We can multiply through by 10 again,so that the number on the L.H.S becomes a whole number.
Therefore, we have
300x = 375 (y)
Put "x=40+y" in the equation above
That is, 300 (40+y) = 375 (y)
1200 + 300y = 375y
1200 = 375y - 300y
1200 = 75y
Divide both sides by 75 to get your y
Therefore, y =16
From equation 1, we had x= 40 + y
Therefore X = 40 + 16
X= 56
Therefore the two unknown numbers are 56 and 16 respectively.
Step-by-step explanation:
Answer:
There is no missing dollar.
Step-by-step explanation:
The question adds twice the $2 kept by the bell boy. Each one paid $9, so $27 of which the room cost 25 and the bellboy kept 2. Each one received 1 dollar back. Room($25)+Bellboy($2)+Change($3)=$30
If we say 27(room)+2(bellboy) you are accounting for the tip twice
<h2>
</h2>
<h3>(OR)</h3>
<h2>
</h2>
![\sf{=\dfrac{d}{dx}\bigg[\dfrac{a^x}{ln(a)} \bigg] }](https://tex.z-dn.net/?f=%5Csf%7B%3D%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbigg%5B%5Cdfrac%7Ba%5Ex%7D%7Bln%28a%29%7D%20%20%5Cbigg%5D%20%7D)
<h3>(Linear differentiation)</h3>
<h2>
</h2>
<h3>✭ Linear differentiation</h3>
→ ![\sf{[a.b(x)+c.d(x)]'=a.b'(x)+c.d'(x)}](https://tex.z-dn.net/?f=%5Csf%7B%5Ba.b%28x%29%2Bc.d%28x%29%5D%27%3Da.b%27%28x%29%2Bc.d%27%28x%29%7D)
<h3>✭ Exponential function rule</h3>
→ 
