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Dafna1 [17]
3 years ago
11

There are two decks of playing cards.

Mathematics
1 answer:
IrinaVladis [17]3 years ago
8 0
A) (1 x 4) + (1 x 4) = 8

There are 8 possible combinations

B) RH:RD:RC:RS:BH:BD:BC:BS


Hope this helps and have a nice day!
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8 batteries cost 10 dollars how much do 6 batteries cost
pychu [463]

6 batteries cost $7.50 because u divide 10 by 8 and then multiply by 6

8 0
3 years ago
Read 2 more answers
Pls help! Asap brainliest to helpful answers
riadik2000 [5.3K]

Answer:

125

Step-by-step explanation:

3 0
3 years ago
Compute the following without a calculator or "guessing and checking." Show all work. (a) What is 7376 mod 23? Hint: Don’t expli
lara31 [8.8K]

Answer:

a. 3(mod 23)

bi d=31 (mod 72)

ii c = 4

iii m = 4

Step-by-step explanation:

Check the attached file for step by step solution

try to open the image in a new tap if the letters appear small

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8 0
3 years ago
Jeremy is having his basketball team over for an end-of-season party. He has $80 to spend on pizza and soda. Pizzas cost $13.99
Lapatulllka [165]

Answer:

13.99p + 4.49s ≤ 80

Step-by-step explanation:

Pizza is p and soda is s.  If he has only $80 to spend on a combination of both, he cannot obviously spend more than that.  Therefore, the inequality sign is "less than or equal to".  He can spend 80, but he can also spend less than 80.  He cannot spend anything over that, since he doesn't have it to spend!

If pizzas are represented by p, and we know the cost of each one is 13.99, we represent one pizza as 13.99p; if the cost of sida is 4.49 per case, we represent one case as 4.49s.  

The sum of these cannot exceed 80, so the inequality, then, is:

13.99p + 4.49s ≤ 80

5 0
4 years ago
Sum of all natural numbers from100to300 which are divisible by 4and 5 is​
My name is Ann [436]

If x is a number that is both divisible by 4 and 5, then

\begin{cases}x\equiv0\pmod4\\x\equiv0\pmod5\end{cases}

4 and 5 are coprime, so we can use the Chinese remainder theorem to solve this system and find that x=20n is a solution to the system, where n is any integer. Simply put, any multiple of 20 fits the bill.

Now, there are 11 numbers between 100 and 300 that are divisible by 20 (100, 120, 140, and so on). We have 20n=100 when n=5, so the sum we want to compute is

\displaystyle\sum_{n=5}^{15}20n=\boxed{2200}

4 0
4 years ago
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