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Flauer [41]
2 years ago
5

at an ice cream store with five flavors of ice cream (peppermint, hoarhound, chocolate malt, gingerbread, and squirrel), ice cre

am scoops are stored inside the ice cream containers. what is the smallest number of ice cream scoops that would need to be in use so that two of the scoops would have to be stored in the same flavor ice cream container? how many ice cream scoops must be in use if 3 of them have to be stored in the same flavor ice cream container?
Mathematics
1 answer:
Darina [25.2K]2 years ago
3 0

ANSWER: 10 ICE-CREAM SCOOPS

There are 5 icecream flavors

For each we need 2 scoops so = 2*5 = 10 scoops

therefore ther must be 10 icecream scoops in use for five flavors.

You can learn more about this thorugh the link below

brainly.com/question/1553688#SPJ4

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kifflom [539]
The first step for solving this problem is to multiply both sides of the bottom equation by -3.
\left \{ {{6x-9y=16} \atop {-6x + 9y = -21}} \right.
Add the two equations together.
6x - 9y - 6x + 9y = 16 - 21
Eliminate the opposites.
-9y + 9y = 16 - 21
Remember that the sum of two opposites equals 0,, so the equation becomes the following:
0 = 16 - 21
Calculate the difference on the right side of the equation.
0 = -5
This means that the statement \left \{ {{6x-9y=16} \atop {2x-3y=7}} \right. is false for any value of x and y. That means that the answer to your question is (x,y) ∈ ∅,, or no solution.
Let me know if you have any further questions.
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8 0
3 years ago
SOMEONE HELP ME NOW. AND I WANT THE RIGHT ANSWER PLEASE
scoray [572]
The first answer is correct
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6 0
3 years ago
Read 2 more answers
Three machines operating independently, simultaneously, and at the same constant rate can fill a certain production order in 36
Andrews [41]

Answer:

D

Step-by-step explanation:

Let the total production order be X. The combined rate is thus x/36 orders per hour.

Now, we know that the three machines are working at the same constant rate. This means that individual rate for each of the machines will be x/36 divided by 3 and that gives x/108 per machine.

Now, we are having another machine coming at the same constant rate. This means we are adding an x/108 rate to the preexisting x/36.

The new total rate thus becomes x/108 + x/36 = 4x/108

Now we know that the total new rate is 4x/108. Since the total work doesn’t change and it is still x, the time taken to complete a work of x orders at a rate of 4x/108 order per hour would be x divided by 4x/108 and this is x * 108/4x = 108/4 = 27 hours

7 0
3 years ago
How to solve a log equation with variable in the exponent?
GREYUIT [131]
Example...

4^x = 16
log(4^x) = log(16)
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6 0
3 years ago
(10 times 2) + 5 times 8 =
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The answer is 60 because 10 x 2 is 20 and 5 x 8 is 40 and add those together to get 60
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