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aksik [14]
4 years ago
15

I need major help please i will give a brainliest to anyone with the correct answer

Mathematics
1 answer:
Allisa [31]4 years ago
8 0
C

hope this helps! good luck. :)
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Is the set of real numbers closed under addition? Explain why or why not. If it is not closed, give an example​
kicyunya [14]

Answer: Yes, the set of real numbers is closed under addition.

Explanation:

Let x and y be two real numbers. Their sum x+y is some other real number. This is what it means when we say the set of real numbers is closed under addition. Taking any two numbers and adding them will get us some other real number.

There is no way to have x+y be nonreal while x,y are both real.

3 0
3 years ago
The ratio of girls to boys is this great is 6 to 8 is there are 120 girls how many boys are there
hram777 [196]

Answer:

160

Step-by-step explanation:

if girls is 6 and boys is 8

and there are 120 girls

120g= 6g * 20

meaning you multiply 8 by the same thing

160= 8b * 20

{\_/}

( >3<) - there

/ > >\

8 0
3 years ago
Find the radius of a circle if the circumference is 66cm​
FrozenT [24]

Answer:

Approximately 10.5, or 10.5095541.

Step-by-step explanation:

To find the radius given the circumference of a circle, we first have to divide the circumference by pi, or an estimation of 3.14. This will give us the diameter.

Since we know that the radius of a circle is simply half the diameter, we divide the number we got from the problem above by 2. This will give you your answer!

7 0
3 years ago
Read 2 more answers
Find how many terms of a geometric progression 1+3+9 are required to make a total of more than 1 million.
Alborosie

The partial sum of a geometric sequence is

\displaystyle \sum_{i=0}^N a^i = \dfrac{a^{N+1}-1}{a-1}

In your case a=3, so if we sum N terms of the sequence we have

\displaystyle \sum_{i=0}^N 3^i = \dfrac{3^{N+1}-1}{2}

We want this to me more than 1 million, so we have

\dfrac{3^{N+1}-1}{2}>1000000 \iff 3^{N+1}-1>2000000 \iff 3^{N+1} > 1999999

Considering the log (base 3) of both sides, we have

N+1>\log_3(1999999)\iff N>\log_3(1999999)-1 approx 12.2

So, starting from N=13, the sum of the first N terms will be more than 1 million

3 0
4 years ago
How do yo solve the problem (2b-7)(8b+6)
trasher [3.6K]
2b-7=8b+6

Combined like terms by subtracting 8b to the other side
8 0
3 years ago
Read 2 more answers
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