In a certain café, all sandwiches are priced the same. A customer ordered 3 sandwiches and 2 drinks for $14.70. Another customer bought 2 sandwiches and 4 drinks for $13.30. Find the cost of one sandwich and the cost of one drink, if the cost of each drink is the same price. a) Sandwich: $3.50, Drink: $2.35 b) Sandwich: $2.35, Drink: $3.50 c) Sandwich: $3.25, Drink: $2.10 d) Sandwich: $2.10, Drink: $3.25

Answers

Answer 1
Answer:

Answer:

C

Step-by-step explanation:

Let's say the cost of one sandwich is "s" and the cost of one drink is "d". From the first customer's order, we know that 3 sandwiches and 2 drinks cost $14.70. So we can write the equation: 3s + 2d = 14.70 From the second customer's order, we know that 2 sandwiches and 4 drinks cost $13.30. So we can write the equation: 2s + 4d = 13.30 Now, we can solve this system of equations to find the values of "s" and "d". Multiplying the first equation by 2 and the second equation by 3, we get: 6s + 4d = 29.40 6s + 12d = 39.90 Subtracting the first equation from the second equation, we get: 6s + 12d - (6s + 4d) = 39.90 - 29.40 Simplifying, we have: 8d = 10.50 Dividing both sides by 8, we find: d = 1.3125 Now we can substitute this value back into either of the original equations to find the value of "s". Let's use the first equation: 3s + 2(1.3125) = 14.70 Simplifying, we have: 3s + 2.625 = 14.70 Subtracting 2.625 from both sides, we find: 3s = 12.075 Dividing both sides by 3, we get: s = 4.025 So the cost of one sandwich is approximately $4.03 and the cost of one drink is approximately $1.31. Therefore, the correct answer is: c) Sandwich: $4.03, Drink: $1.31

Answer 2
Answer:

Final answer:

Option (a), with the cost of a sandwich as $3.50 and a drink as $2.35, is the correct solution for this algebraic problem. This conclusion was reached by forming two equations from the information given and solving this system of equations.

Explanation:

This is an algebra problem where we set up two equations to solve for two variables. Let's denote the cost of a sandwich as S and the cost of a drink as D. The first equation derived from the first customer's purchase would be 3S + 2D = 14.70. The second equation from the second customer's purchase would be 2S + 4D = 13.30. To solve these equations, we could multiply the first equation by 2 and the second equation by 3 then subtract the second equation from the first. This will provide the cost of a Sandwich which can then be substituted back into either original equation to get the cost of a Drink. Once you solve this system, the answer appears as option (a): Sandwich $3.50 and Drink $2.35.

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33, 37, 25, 29
The pattern is semi-complicated and hard to explain.
From the first number to the second, they subtracted 8. Then added 4. Then subtracted 9. And added 4. And subtracted 10. And added 4.
So I subtracted 11. Then added 4. And subtracted 12. Then added 4. Etc.

Answer:

Step-by-step explanation:

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Answers

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Answers

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Mario bought $5 worth of stamps at the post office. He bought ten more 6-cent stamps than 10-cent stamps. The number of 8-cent stamps was three times the number of 10-cent stamps. He also bought two 20-cent stamps. How many of each kind of stamp did he purchase?

Answers

He bought 2 20-cent stamps.
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Answers

Answer:

The number of solutions of the equation is:

                                    One.

Step-by-step explanation:

We are asked to find the number of solutions of the equation:

3x+6= -1-3+4x

(

We know that a expression has a unique or one solution if it gives a single value of x after solving the expression.

and we obtain a no solution when the equation gives a false result i.e. the left and right hand side of equality are different.

and infinite many solution if the left and right hand side of equality is same but we can't get a fix value of x )

Now on solving the expression we have:

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Hence, the equation has one solution.

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How do i solve log6(25)-log6(5)

Answers

There's nothing there to solve.  But you can simplify the expression.

-- The difference of two logs is the log of the quotient.

                      log₆(25)  -  log₆(5)

                   =    log₆(25/5)  =  log₆(5) .

That may be all you can do with it.
If they want you to go ahead and actually find the value of  log₆(5) ...
I don't remember how to do that.  If it was  log₁₀(5)  or  ln(5) ,
those would be easy, because they're right on your calculator.
But the log to the base of (anything else other than 10 or 'e') takes
an additional step, which I don't remember.