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BlackZzzverrR [31]
4 years ago
15

How many 1 inch cubes do you need to create a cube with an edge length of 5 inches

SAT
1 answer:
Gala2k [10]4 years ago
4 0

Answer:

You would need 125 1-inch cubes

Step-by-Step Explanation:

A cube with an edge length of 5 has a volume of 125 cubic inches (5x5x5 equals 125). A cube with an edge length of 1 has a volume of 1 cubic inch (1x1x1 equals 1). Divide 125 by 1 and you will find that 125 1-inch cubes are required to create cube with an edge length of 5.

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for making a scatter plot of the data set, which would be the best scale and interval for the axis that represents the amount of
Sphinxa [80]

The best scale and interval to adopt for the amount of Fluoride axis would be 0 to 8 with intervals of 0.2

To make a decision on the scale and interval to adopt when making a graph ;

  • We need to examine the the range of the values in the distribution.

  • Data for the amount of Fluoride ranges from 0.4 to 7.1

  • The number of values in the distribution is 7.

  • Lower scale value must be below the minimum Fluoride value preferably 0

  • Maximum scale value must be above the maximum Fluoride value.

  • Hence an interval of 0 to 8 would suit the axis.

  • The interval is the value of each unit represented on the graph.

  • A scale of 4 would be too large to depict all the 7 values with an interval of 8 units

  • Hence, the most reasonable scale from the option would be 0.2 as it would allow the points to be properly spaced on the graph.

Learn more : brainly.com/question/1846548

7 0
2 years ago
Explain what the notation lim f(x)=-infinity means
mote1985 [20]

Answer:

You are finding the horizontal asyemtopes/end behavior of the graph.

Explanation:

You are finding the horizontal asyemtopes/end behavior of the graph.

5 0
3 years ago
Ry owns a commercial building and leased his premises to John to conduct a bakery business. Under the lease contract, John is ob
Nata [24]

Answer:

See explaination

Explanation:

The Issue : Gary being the lessor that leased his commercial property to John to conduct bakery business with an obligation to repair modified or damaged parts if any after lease term. However, John couldnt repair the same but paid $3,100 to Gary to cover the repair expenses.

The Legislation : The required Provision says that lessee is reponsible to keep the premises in good condition and to handover back to the lessor in the same condition as it was in the beginning of the lease period except normal wear and tear. Here damage to the building was not due to passage of time or Act of God . It was due to installation of Machinery and fixtures in the building by the lessee.

According to Polster, Inc. v. Swing, tenant was reponsible to repair the damage to the drop ceiling, ceiling tiles, interior walls, front door sill and jamb as it was not normal wear and tear.

Also in the case of Churchill Forge, Inc. v. Brown, there is no unjust in requiring a tenant to reimburse the expenses to landlord. As a result cost will shift to the tenant/lessee as the case may be.

In Conclusion : Thus it is the responsibilty of John to repair the property or else to reimburse the entire expenses incurred to Gary to put the building in good shape other than for normal wear and tear.

Here , the installing of machinery and fixtures should be considered as capital expenditure as lonterm benefit is determined. However, repairs at the end of the lease period will be categorised as revenue expenditure and considered in Profit & loss account for the purpose of tax.

4 0
3 years ago
Which mathematical word describes both 27 x and 6 x in the expression 27 x + 6 x ?
dangina [55]
Commutative is it i think
4 0
2 years ago
Suppose that 3. 33 g of acetone at 25. 0 °c condenses on the surface of a 44. 0-g block of aluminum that is initially at 25 °c.
Vlada [557]

The final temperature on the block is 25 degrees.

Heat is a form of energy. This question shows that energy is transferred from the acetone to the metal block.

Data given;

  • M1 = 3.33g
  • T1 = 25 degrees
  • M2 = 44g
  • T2 = 25 degree
  • C1 = 124.5 J mol K
  • C2 = 0.89 J/g K
<h3>Heat Transfer</h3>

Applying the principles of heat transfer,

Heat loss = Heat gain

m1c1(T1 - T3) = m2c2(T3 - T2)\\3.33*124.5*(25.0 - T3) = 44*0.89*(T3 - 25)\\10364.625 - 414.585T3 = 39.16T3 - 979\\10364.625+979 = 414.585T3 +39.16T3\\11343.25 = 453.745T3\\T3 = \frac{11343.25}{453.745} \\T3 = 24.999\\T3 = 25

From the calculation above, the final temperature on the block is 25 degrees.

Learn more on heat transfer here;

brainly.com/question/15003174

3 0
2 years ago
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