The equilibrium price is $12.
<h3>What is the
equilibrium price ?</h3>
Equilibrium is the point where the quantity demanded is equal to the quantity supplied. The price at equilibrium is known as the equilibrium price and the quantity at equilibrium is known as the equilibrium quantity.
When shown on a graph, equilibrium is the point where the quantity demanded curve is equal to the quantity supplied curve.
When there is equilibrium, the equation of quantity demanded would be equal to the equation of quantity supplied.
-280 + 40p = 800 - 50p
In order to determine the value of p, take the following steps:
Combine similar terms: 800 + 280 = 40p + 50p
Add similar terms = 1080 = 90p
Divide both sides of the equation by 90 : 1080 / 90 = 12
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Answer:
± 27.33 ft
Step-by-step explanation:
For the given problem, we can estimate the initial and final coordinates of the line of the ball path as (-40,-50) and (0,0). Therefore, the slope is:
(-50-0)/(-40-0) = 50/40 = 1.25
Similarly, we can estimate the slope of a perpendicular line to the line of the ball path as: -1*(1/1.25) = -0.8.
Therefore, using (0,0) and the slope -0.8, the equation of the perpendicular line is: -0.8 = (y-0)/(x-0);
-0.8 = y/x
y = -0.8x
Furthermore, we are given the circle radius as 35 ft and we can use the distance formula to find the two points 35 ft far from the origin:
35^2 = x^2 + y^2
y = -0.8x
35^2 = x^2 + (-0.8x)^2
1225 = (x^2 + 0.64x^2)
1225 = 1.64x^2
x^2 = 1225/1.64 = 746.95
x = sqrt(746.95) = ± 27.33 ft
Answer:
The correct answer is d
Step-by-step explanation:
Answer:
y = m x + b equation of a straight line
m m' = -1 condition for perpendicular lines
If y = 4 x - 7 then m = 4 so m' = -.25
Y = -.25 X + A we need to find A
A = Y + .25 * 8 = 2 + 2 = 4
Y = -.25 X + 4
Check:
2 = -.25 * 8 + 4 = -2 + 4 = 2
<h2>
Greetings!</h2>
Answer:
6a + 1
Step-by-step explanation:
First, you need to multiply everything in the brackets by 3L
(7 + 2a) x 3 =
7 * 3 = 21
2a * 3 = 6a
21 + 6a
Now you can subtract the 20 from this:
6a + 21 - 20
Simplified down:
6a + 1
<h2>Hope this helps!</h2>
