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likoan [24]
3 years ago
5

The density of a certain material is such that it weighs 71 pounds for every 2.5 gallons of volume. Express this density in tons

per cubic meter. Round your answer to the nearest hundredth.
Mathematics
1 answer:
kvasek [131]3 years ago
5 0

Answer:

d=3.75\ \text{tons per cubic meter}

Step-by-step explanation:

Given that,

Mass of a certain material, m = 71 pounds

Since, 1 ton = 2000 pounds

71 pounds = 0.0355 ton

Volume of the material, V = 2.5 gallons

Since, 1 m³ = 264.172 gallon

2.5 gallon = (2.5/264.172 ) m³

Density,

d=\dfrac{m}{V}\\\\d=\dfrac{0.0355\ \text{tons}}{\dfrac{2.5}{264.172}\ m^3}\\\\d=3.75\ \text{tons per cubic meter}

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3 years ago
3/5 = 12/x find the missing numbers in the equivalent ratios
pogonyaev

Answer:

x=20

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2 years ago
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Country days scholarship rounds receive a gift of $135000. The money is invested in stock, bonds, and CDs. CDs pay 2.75% interes
Vsevolod [243]

Answer:

  • CDs — $10,000
  • bonds — $80,000
  • stocks — $45,000

Step-by-step explanation:

Let the variables c, b, s represent the dollar amounts invested in CDs, stocks, and bonds, respectively. Then the problem statement gives us 3 relations between these 3 variables:

  c + b + s = 135000 . . . . . . . . . . . . . . . . . total invested

  0.0275c +.045b +0.104s = 8555 . . . . . total income earned

  -c + b = 70000 . . . . . . . . . . . . . . . . . . . . . 70,000 more was in bonds than CDs

Using the third equation to write an expression for b, we can substitute into the other two equations.

  b = 70000 +c . . . . . . . . . . . . . . . . expression we can substitute for b

  c + (70000 +c) +s = 135000 . . . . substitute for b in the first equation

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  .0275c +.045(70000 +c) +.104s = 8555 . . . . . substitute for b in 2nd eqn

  .0725c +.104s = 5405 . . . . . . . . . . [eq5] simplify

Using [eq4], we can write an expression for s that can be substituted into [eq5].

  s = 65000 -2c . . . . . . . expression we can substitute for s

  0.0725c +0.104(65000 -2c) = 5405

  -0.1355c = -1355 . . . . . . . . . . . . . . . . . . . . subtract 6760, simplify

  c = 1355/.1355 = 10,000

  s = 65000 -2×10000 = 45,000

  b = 70000 +10000 = 80,000

The amounts invested in stocks, bonds, and CDs were $45,000, $80,000, and $10,000, respectively.

_____

Alternatively, you can reduce the augmented matrix for this problem to row-echelon form using any of several calculators or on-line sites. That matrix is ...

\left[\begin{array}{ccc|c}1&1&1&135000\\0.0275&0.045&0.104&8555\\-1&1&0&70000\end{array}\right]

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3 years ago
A new roll of wrapping paper has 658 yards of paper. drew used 216 yards from the roll to wrap a gift. which equation most close
igor_vitrenko [27]
658-216=442
There should be 442 yards of wrapping paper left on the roll.
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See the attached picture.

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