Answer:
He reached the top of the tree on the 31st day
Step-by-step explanation:
The total distance for the owl to climb is 93 units.
If the owl climbed 18 units up the tree every day without coming down, the owl would have taken 93/18 days to reach the top of the tree.
However, the owl descends by 15 units every night. This just reduces the overall distance covered at the end of each day's climb.
The net distance covered by the owl after each day is 18-15 units = 3 units of climb. This is the steady distance the owl gains up the three at the end of each day after the ascent and descent.
The time taken for the climb moving at this pace of 3 units per day will be
93 units / 3 units per day = 31 days
Answer:
If you have given an equation, you see the x and y in it. Just taking second equation and think any common factor between x's variable. With common factor, multiply both and you will find the value of y and put it in any equation, you will find the value of X.
Step-by-step explanation:
That is highest to lowest value of y
we know the range of y=cos x is -1 to 1
so
we multiply it by 3
-3 to 3 is the range
X =126 because if you multiply 9 and 14 you get 126 and 126 divided by 9 is 14
Answer:
$7812.5million
$7812.5million
Step-by-step explanation:
From the question, the demand quantity <em>D(x) is 410-x </em> and the supply quantity <em>S(x)=160+x.</em> We determine the equilibrium quantity by equating the Demand quantity and the Supply quantity i.e
<em>D(x)=S(x). </em>
<em>
</em>
Hence the equilibrium quantity is 125.
Next we determine the equilibrium price. This can be obtain by just substituting the equilibrium quantity into either the demand quantity or the supply quantity. I prefer using the Demand quantity.
Equilibrium price=
.
Next we write the expression for the Consumer Surplus
where <em>p</em> and <em>q </em> are the equilibrium price and equilibrium quantity respectively.
By substituting values we have
By carrying out simple arithmetic we arrive at
.
To determine the producer surplus, we use the expression below
Hence if we substitute values we arrive at
![PS=(285)(125)-\int\limits^q_0 {(160+x)} \,dx\\PS=(285)(125)-[(160x-\frac{x^{2} }{2})]^{125}_{0}](https://tex.z-dn.net/?f=PS%3D%28285%29%28125%29-%5Cint%5Climits%5Eq_0%20%7B%28160%2Bx%29%7D%20%5C%2Cdx%5C%5CPS%3D%28285%29%28125%29-%5B%28160x-%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B2%7D%29%5D%5E%7B125%7D_%7B0%7D)
.
By simply simplification we arrive at
