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NikAS [45]
2 years ago
7

Help??? Please math image

Mathematics
1 answer:
Olenka [21]2 years ago
4 0

Answer:

area: 24

lateral area: 36

total area: I am not sure.

volume: 24

Step-by-step explanation:

area: length times width

lateral area: (perimeter of base) times height

total area: I do not know, sorry.

volume: length times width times height.

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On a​ map, 1 inch equals 10 miles. Two cities are 8 inches apart on the map. What is the actual distance between the​ cities?
cestrela7 [59]

Answer:

80 inches the reason for this is 10 times 8 equals 80

5 0
2 years ago
Which is NOT a common denominator for 2/3 and 1/2 ?<br><br> A. 18<br> B. 12<br> C. 6<br> D. 3
Iteru [2.4K]

Answer:

d. 3

Step-by-step explanation:

hope this helps!!!!!!!<3

3 0
3 years ago
Read 2 more answers
What is the first step in solving this system by elimination?
kherson [118]

You want to try and get rid of one of the variables ( x or y) so you would multiply the second equation by 2 which would make the x variable -2x which then when added to the first equation would eliminate the x variable.

Answer: c. Multiply the second equation by 2.

4 0
3 years ago
Si mi disco duro tiene una capacidad de 1.5 Terabytes ¿ cuántas llaves de 32 Gigabits caben en ese disco duro?
Art [367]

Answer:

The number of 32 Gigabit keys that can be fitted on the hard drive is 375.

Step-by-step explanation:

The question is:

If my hard drive has a capacity of 1.5 Terabytes, how many 32 Gigabit keys can fit on that hard drive?

Solution:

1 Terabyte = 8000 Gigabits

Then 1.5 Terabytes in Gigabits is:

1.5 Terabytes = (8000 × 1.5) Gigabits

                      = 12000 Gigabits

One key is of 32 Gigabits.

Compute the number of 32 Gigabit keys that can be fitted on the hard drive as follows:

\text{Number of 32 Gigabit keys}=\frac{12000}{32}=375

Thus, the number of 32 Gigabit keys that can be fitted on the hard drive is 375.

4 0
3 years ago
Increments of seven ​
BaLLatris [955]

Answer:

Specific Learning Outcomes:  

Solve problems that involve finding powers of a number

Description of mathematics:  

In this problem students work with powers of numbers and, as a consequence, come to understand what is happening to the numbers.  

Students also see how an apparently enormous and difficult calculation can be broken down into manageable parts. The students should come to realise that there are only a limited number of unit digits obtained when 7 is raised to a power. Further,  these specific digits 'cycle round' as the power of 7 increases. This cycle is 7, 9, 3, 1, 7, 9, …

The same is true of the digit in the tens place.

6 0
3 years ago
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