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Katen [24]
3 years ago
14

Find three consecutive integers whose sum is 21

Mathematics
2 answers:
stira [4]3 years ago
5 0

Answer:

6 + 7 + 8 = 21

Step-by-step explanation:

6, 7, and 8 are consecutive integers.

6 + 7 is 13. 13 + 8 = 21. So, 6 + 7 + 8 are consecutive integers that equal 21.

Flauer [41]3 years ago
3 0

Answer:

6+7+8= 21

Step-by-step explanation:

6,7,8 follow consecutively. and if you add them, the sum is 21

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How many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code?
ad-work [718]
Note that powers of 2 can be written in binary as

2^0=1_2
2^1=10_2
2^2=100_2

and so on. Observe that n+1 digits are required to represent the n-th power of 2 in binary.

Also observe that

\log_2(2^n)=n\log_22=n

so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,

2_{10}=10_2\iff\log_2(2^1)=\log_22=1
3_{10}=11_2\iff\log_23=1+(\text{some number between 0 and 1})
4_{10}=100_2\iff\log_24=2

That is, both 2 and 3 require only two binary digits, so we don't care about the decimal part of \log_23. We only need the integer part, \lfloor\log_23\rfloor, then we add 1.

Now, 2^9=512, and 999 falls between these consecutive powers of 2. That means

\log_2999=9+\text{(some number between 0 and 1})

which means 999 requires \lfloor\log_2999\rfloor+1=9+1=10 binary digits.

Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write 1_2? Or do you pad this number with 0s to get 10 digits, i.e. 0000000001_2? In the latter case, the answer is obvious; 1000\times10=10^4 total binary digits are needed.

In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.
8 0
3 years ago
In the diagram shown, lines m and n are
bonufazy [111]

Answer:

Step-by-step explanation:

If two line 'm' and 'n' are parallel and a transversal line 't' i intersecting these lines at two distinct points,

Sum of interior consecutive angles should be 180°.

(Property of the angles formed between parallel lines and a transversal line)

Therefore, 114° + 64° = 180°

178° = 180°

Which is not true.

Therefore, lines 'm' and 'n' are not the parallel lines.

3 0
3 years ago
3 octagons 4 Pentagons 2 trapezoids how many sides are on the shapes altogether
klio [65]

Answer:

52 Sides

Step-by-step explanation:

Octagons have 8 sides so 3 x 8 = 24

Pentagons have 5 sides so 4 x 5 = 20

Trapezoids have 4 sides so 2 x 4 = 8

Add it all up and it equals 52 sides in total

8 0
3 years ago
[BRAINLIEST] a. Nick deposits money in a Certificate of Deposit account (CD). The balance (in dollars) in his account t years af
crimeas [40]

Answer:

(1.06)0 = 1  and positive powers of 1.06 are larger than 1, thus the minimum value N(t) attains, if t≥0, is 400.

From the point of view of the context, a CD account grows in value over time so with a deposit of $400 the value will never drop to $399.

4 0
2 years ago
Which of the following numbers is both a perfect square and a perfect cube?
Marat540 [252]

Answer:

531441

Step-by-step explanation:

We can find by   factorization

531441 = 9 * 9 * 9 * 9 * 9 *9 = (9*9*9) * (9*9*9)

\sqrt[3]{531441}=\sqrt[3]{9*9*9*9*9*9}  =9*9 = 81\\\\\\\sqrt{531441}=\sqrt{(9*9)*(9*9)*(9*9)}=9*9*9=729

7 0
3 years ago
Read 2 more answers
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