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qwelly [4]
3 years ago
14

Write a word problem for the inequality -3.5 < -2.5

Mathematics
1 answer:
nlexa [21]3 years ago
8 0
Jack is in debt after borrowing some money. He has more than -3.5 dollars and less than -2.5 dollars. Write this as an inequality.
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Help plz math question. EXPLAIN HOW YOU GOT THE ANSWER. 30 PTS btw thx if u help
pentagon [3]

Answer:

256

Step-by-step explanation:

4^4^4 = 4^(4^4) = 4^256

so written in base 4, there will be 256 zeroes after a 1.

This translates to approximately 154 digits in decimal.

Just like 2^2^2 =  2^4

there will be four zeroes after a 1.

2^2^2=2^(2^2) = 2^4 = 16 = 10000 (16 in base 2).

4 0
3 years ago
How do you find the least common for 3 and 8?
sesenic [268]

The least common multiple of 3 and 8 is 24.

The least common multiple of two numbers is the lowest value number that is a multiple of both the numbers. To find the least common multiple of 3 and 8, you first list their multiples:

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24...

Multiples of 8: 8, 16, 24, 32, 40, 48, 56...

When you compare the two lists of multiples, you can see that the number 24 is the lowest number that is a multiple of both 3 and 8, so it is their LCM, or least common multiple.


<em>HOPE THAT HELPS!!!!</em>

8 0
3 years ago
I did not do as good as I hoped on my math model :(
pochemuha
I’m sorry! I hope you do better next time.
8 0
3 years ago
Read 2 more answers
When do we need to consider amounts that do not represent whole numbers?
Alexandra [31]
When we are dealing with fractions and decimals.
4 0
3 years ago
I need some help finding the volume of a sphere with a diameter of 18.6cm. The website says the answer I got from my calculator
lutik1710 [3]

Answer:

V ≈ 3367.6 cm³

Step-by-step explanation:

The volume (V) of a sphere is calculated as

V = \frac{4}{3} πr³ ( r is the radius )

Here D = 18.6 , then r = 18.6 ÷ 2 = 9.3 , so

V = \frac{4}{3} × 3.14 × 9.3³

   = \frac{4}{3} × 3.14 × 804.357

   = \frac{4(3.14)(804.357)}{3}

   ≈ 3367.6 cm³ ( to the nearest tenth )

6 0
2 years ago
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