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Margaret [11]
4 years ago
11

Although the actual amount varies by season and time of​ day, the average volume of water that flows over the falls each second

is 5.25.2 times ×10 Superscript 5105 gallons. How much water flows over the falls in an​ hour? Write the result in scientific notation.​ (Hint: 1 hour equals 3600​ seconds)
Mathematics
1 answer:
Troyanec [42]4 years ago
4 0

The amount of water flowing each second is:

rate = 5.2 x 10^5 gallons / second

 

Since we know that:

1 hour = 3600 seconds

 

Therefore:

rate = (5.2 x 10^5 gallons / second) * (3600 seconds / hour)

rate = 1.872 x 10^9 gallons / hour

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Answer:x<-18

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Step 1: Simplify both sides of the inequality.

-1/2x+3>12

Step 2: Subtract 3 from both sides.

-1/2x+3-3>12-3

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Step 3: Multiply both sides by 2/(-1).

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3 years ago
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Ask Your Teacher The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm. (a) Use differentials t
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Answer:  

a)  28,662 cm²  max error

    0,0111     relative error

b) 102,692 cm³  max error

   0,004     relative error

   

Step-by-step explanation:

Length of cicumference is: 90 cm

L = 2*π*r

Applying differentiation on both sides f the equation

dL  =  2*π* dr    ⇒  dr = 0,5 / 2*π

dr =  1/4π

The equation for the volume of the sphere is  

V(s) =  4/3*π*r³     and for the surface area is

S(s) = 4*π*r²

Differentiating

a) dS(s)  =  4*2*π*r* dr    ⇒  where  2*π*r = L = 90

Then    

dS(s)  =  4*90 (1/4*π)

dS(s) = 28.662 cm²   ( Maximum error since dr = (1/4π) is maximum error

For relative error

DS´(s)  =  (90/π) / 4*π*r²

DS´(s)  = 90 / 4*π*(L/2*π)²      ⇒   DS(s)  = 2 /180

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b) V(s) = 4/3*π*r³

Differentiating we get:

DV(s) =  4*π*r² dr

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DV(s)  =  102,692 cm³   max error

Relative error

DV´(v) =  (90)² / 8*π²/ 4/3*π*r³

DV´(v) = 1/240

DV´(v) =  0,004

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Just add 360 to -800 until you end up with an angle between 0 and 360:
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So the angle is 280 degrees.
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