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IrinaVladis [17]
3 years ago
14

Six friends went out to lunch they each started with the same amount of money and they each spent 8$ they ended with a combined

total of 48$ how much money did each of them have to start
Mathematics
1 answer:
ss7ja [257]3 years ago
4 0

Answer:

8

Step-by-step explanation:

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2 decreased by the sum of m and 8
Elina [12.6K]

Answer:

(m+8) - 2

Step-by-step explanation:

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2 years ago
Use Fermat's Little Theorem to determine 7^542 mod 13.
m_a_m_a [10]

a^{p-1} \equiv 1 \pmod p where p is prime, a\in\mathbb{Z} and a is not divisible by p.

7^{13-1}\equiv 1 \pmod {13}\\7^{12}\equiv 1 \pmod {13}\\\\542=45\cdot12+2\\\\7^{45\cdot 12}\equiv 1 \pmod {13}\\7^{45\cdot 12+2}\equiv 7^2 \pmod {13}\\7^{542}\equiv 49 \pmod{13}

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3 years ago
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Name the postulate or theorem, 50 points please help!
max2010maxim [7]
Hi! 

When you use postulates and theorems, you need to make sure to only use the given information that you know. Look for the given statements, and congruence marks on the figure. Those are also considered given. 

By looking, you are given an angle and a side. The side comes first. SU=TV.

So, that makes it so Side-Angle-Side would be the best option. 

I hope this helps! 


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3 years ago
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On her pogo stick lula made 24 hops in 30 seconds at this rate how many hops in 50 seconds
Lorico [155]
44 hops in 50 seconds
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3 years ago
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A concert hall has 8,000 seats and two categories of ticket prices, $29 and $34. Assume that all seats in each category can be s
Greeley [361]

Using a system of equations, it is found that:

  • For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.
  • For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.
  • For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.

--------------------

The variables of the system are:

  • x, which is the number of $29 tickets sold.
  • y, which is the number of $34 tickets sold.

Total of 8,000 seats, all can be sold, thus:

x + y = 8000 \rightarrow x = 8000 - y

--------------------

For a profit of $255,000, we have that:

29x + 34y = 255000

29(8000 - y) + 34y = 255000

5y = 23000

y = \frac{23000}{5}

y = 4600

x = 8000 - 4600 = 3400

For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.

--------------------

For a profit of $271,000, we have that:

29x + 34y = 271000

29(8000 - y) + 34y = 271000

5y = 39000

y = \frac{39000}{5}

y = 7800

x = 8000 - 7800= 200

For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.

--------------------

For a profit of $235,000, we have that:

29x + 34y = 235000

29(8000 - y) + 34y = 235000

5y = 3000

y = \frac{3000}{5}

y = 600

x = 8000 - 600 = 7400

For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.

A similar problem is given at brainly.com/question/22826010

3 0
3 years ago
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