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Luba_88 [7]
3 years ago
9

Which is more, 5 tons or 9,998 pounds?

Mathematics
2 answers:
Katyanochek1 [597]3 years ago
7 0

Answer:

5 tons

Step-by-step explanation:

Flauer [41]3 years ago
5 0
5 tons because 5 tons is 10,000 pounds
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Over a certain region of space, the electric potential is V = 4x − 7x2y + 7yz2. Find the expressions for the x, y, z components
Tju [1.3M]

Answer:

Ex = - 4 + 14xy

Ey = 7x² - 7z²

Ez = -14yz

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The relationship between Electric field Er(x, y, z) and Electric potential, V, is a differential relationship:

Er(x, y, z) = -dV/dr(x, y, z)

Where r(x, y, z) = distance in x, y and z components.

The x component of the electric field is:

Ex = -dV/dx

Given that:

V = 4x - 7x²y + 7yz²

Ex = -dV/dx

Ex = -(4 - 14xy)

Ex = -4 + 14xy

The y component of the electric field is:

Ey = -dV/dy

Ey = -(-7x² + 7yz²)

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Ez = -14yz

8 0
3 years ago
10. Three kinds of teas are worth $4.60 per pound, $5.75 per pound, and $6.50 per pound. They are to be
zepelin [54]

Answer:

The mass of the $4.60/lb tea that should be used in the mixture is 10 lb

The mass of the $5.75/lb tea that should be used in the mixture is 8 lb

The mass of the $6.50/lb tea that should be used in the mixture is 2 lb

Step-by-step explanation:

The parameters of the question are;

The worth of the three teas are

Tea A = $4.60/lb

Tea B = $5.75/lb

Tea C = $6.50/lb

The mass of the mixture of the three teas = 20 lb

The worth of the mixture of the three teas = $5.25 per pound = $5.25/lb

The amount of the $4.60 in the mixture = The sum of the amount of the other two teas

Therefore, given that the mass of the mixture = 20 lb, we have in the mixture;

The mass of tea A + The mass of Tea B + The mass of Tea C = 20 lb

The mass of tea A = The mass of Tea B + The mass of Tea C

Therefore;

The mass of tea A + The mass of tea A = 20 lb

2 × The mass of tea A in the mixture = 20 lb

The mass of tea A in the mixture = 20 lb/2 = 10 lb

The mass of tea A in the mixture = 10 lb

The mass of Tea B + The mass of Tea C = The mass of tea A = 10 lb

The mass of Tea B + The mass of Tea C = 10 lb

The mass of Tea B  = 10 lb - The mass of Tea C

Where the mass of Tea C in the mixture = x, we have;

The mass of Tea B in the mixture = 10 lb - x

The cost of the 10 lb of tea A = 10 × $4.60 = $46.0

The worth of the tea mixture = 20 × $5.25 = $105

The worth of the remaining 10 lb of the mixture comprising of tea A and tea B is given as follows;

The worth of Tea B + The worth of Tea C in the mixture = $105.00 - $46.00 = $59.00

Therefore, we have;

x lb × $6.50/lb + (10 - x) lb × $5.75/lb = $59.00

x × $6.50 - x × $5.75 + $57.50 = $59.00

x × $0.75 = $59.00 -  $57.50 = $1.50

x =  $1.50/$0.75 = 2 lb

∴ The mass of Tea C in the mixture = 2 lb

The mass of Tea B in the mixture = 10 lb - x = 10 lb - 2 lb = 8 lb

The mass of Tea B in the mixture = 8 lb

Therefore, since we have;

Tea A = $4.60/lb

Tea B = $5.75/lb

Tea C = $6.50/lb

The mass of tea A in the mixture = 10 lb

The mass of tea B in the mixture = 8 lb

The mass of tea C in the mixture = 2 lb, we find;

The mass of the $4.60/lb tea that should be used in the mixture = 10 lb

The mass of the $5.75/lb tea that should be used in the mixture = 8 lb

The mass of the $6.50/lb tea that should be used in the mixture = 2 lb.

6 0
3 years ago
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