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dangina [55]
3 years ago
14

On a vacation trip, the smith family traveled 2,648 miles by car to a lake resort on the return trip they took a shorter route a

nd drove 2,023 miles at the end of the trip the car’s odometer read 46,238 miles what did the odometer read at the beginning of the trip
Mathematics
1 answer:
irina1246 [14]3 years ago
3 0

Answer: 41,567

Step-by-step explanation:

1st: Add 2,648 + 2,023 and get = 4,671

2nd: Subtract 46,238 - 4,671 = 41,567

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Kisachek [45]
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Suppose the supply function for a certain item is given by S(q)= (q+6)2 and the demand funtion is given by D(q)= (1000)/(q+6).
nasty-shy [4]

Answer: The equilibrium point is where; Quantity supplied = 100 and Quantity demanded = 100

Step-by-step explanation: The equilibrium point on a demand and supply graph is the point at which demand equals supply. Better put, it is the point where the demand curve intersects the supply curve.

The supply function is given as

S(q) = (q + 6)^2

The demand function is given as

D(q) = 1000/(q + 6)

The equilibrium point therefore would be derived as

(q + 6)^2 = 1000/(q + 6)

Cross multiply and you have

(q + 6)^2 x (q + 6) = 1000

(q + 6 )^3 = 1000

Add the cube root sign to both sides of the equation

q + 6 = 10

Subtract 6 from both sides of the equation

q = 4

Therefore when q = 4, supply would be

S(q) = (4 + 6)^2

S(q) = 10^2

S(q) = 100

Also when q = 4, demand would be

D(q) = 1000/(4 + 6)

D(q) = 1000/10

D(q) = 100

Hence at the point of equilibrium the quantity demanded and quantity supplied would be 100 units.

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3 years ago
What time is 46 minutes before 2:10 pm
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Find dy du , du dx , and dy dx .
dangina [55]

Answer:

a) \frac{dy}{du}=5(x^2 +1)^4, \frac{du}{dx}=2x, \frac{dy}{dx}=10x(x^2+1)^4  

b) \frac{dy}{du}=4(4x^2-x+6)^3, \frac{du}{dx}=8x-1, \frac{dy}{dx}=4(8x-1)(4x^2-2+6)^3  

Step-by-step explanation:

We can use the chain rule in the following form: is u=u(x) is a differentiable function depending on x and y=y(u) is a differentiable function depending on u, then \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}.

a) \frac{dy}{du}=\frac{d}{du} (u^5)=5u^4=5(x^2 +1)^4 from the power rule.  

\frac{du}{dx}=\frac{d}{dx} (x^2 +1)=2x.

From the previous parts and the chain rule, \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}=5(x^2 +1)^4(2x)=10x(x^2+1)^4  

b) \frac{dy}{du}=\frac{d}{du} (u^4)=4u^3=4(4x^2-x+6)^3  

\frac{du}{dx}=\frac{d}{dx} (4x^2-x+6)=8x-1 from the power and sum rules.

Then, \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}=4(4x^2 -x+6)^3(8x-1)=4(8x-1)(4x^2-x+6)^3  

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