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lys-0071 [83]
3 years ago
7

A keycode must contain 2 letters and 3 numbers. The letters may be any letter of the alphabet. The numbers should be any number

from 0 to 9. How many different keycode combinations are there?
Mathematics
2 answers:
Kitty [74]3 years ago
6 0
It’s actually 676,000. That was the correct answer because I got it wrong on apex.
FromTheMoon [43]3 years ago
3 0
They're 26 Letters in the Alphabet so you do 26x26=676
You can only Pick from 0 to 9
So you would Multiply 10x10x9=900
Add 900 + 676= 1576 They're 1576 Different KeyCode Combinations.
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Lena won a charity raffle. Her prize will be randomly selected from the 9 prizes shown below. The prizes include 7 rings, 1 came
pickupchik [31]

Answer:

(b) Find the odds in favor of Lena winning a headset ​

Step-by-step explanation:

4 0
3 years ago
There are 5 kids who want to start a game of Red Rover. Then 12 more kids come to play. If there are 2 teams in Red Rover, which
mixas84 [53]

Answer: The answer is one will have 8 and one will have 9 so roughly 8

Step-by-step explanation

So to start we need to add all the kids together that are playing

5+12= 17

Now we need to divide the number of kids by the number of teams that there are

17 divided by 2=8.5

So each team will have 8 but one will have nine

Hope this helps :D

7 0
3 years ago
Read 2 more answers
Need help ASAP!!!!! I WILL GIVE THE MOST BRAINLIEST TO THE FIRST PERSON THAT ANSWERS
lys-0071 [83]

Answer:

x^2 + 10x + y^2 -4y + -7 = 0

Step-by-step explanation:

For any circle point with coordinate of center (a,b) and radius r

equation of circle is given by

(x-a)^2 + (y-b)^2 = r^2

_______________________________________

Given

coordinate of center =  (-5,2)

radius = 6

thus, equation of circle can be written by substituting a and b with (-5 ,2) and r =6 as shown below

(x-a)^2 + (y-b)^2 = r^2\\(x-(-5))^2 + (y-2)^2 = 6^2\\x^2 + 10x + 25 + y^2 -4y + 4 = 36\\=>x^2 + 10x + y^2 -4y + 29 = 36\\=>x^2 + 10x + y^2 -4y + 29 -36 = 36 -36\\=>x^2 + 10x + y^2 -4y + -7 = 0

Thus, equation of circle is x^2 + 10x + y^2 -4y + -7 = 0

3 0
3 years ago
Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

6 0
4 years ago
How to Simplify 6y + 5z + 8y – 3z?
Gekata [30.6K]
To simplify, you just condense all of the values with the same variable, based on the signs of those values.
In this example,
6y + 5z + 8y - 3z
is the same as 
6y + 8y + 5z - 3z
which can be simplified to 
14y + 2z
which is the answer to your question.
Hope that helped =)
4 0
3 years ago
Read 2 more answers
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