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Vesnalui [34]
3 years ago
15

There are 15 tables in camille's cafe. some of those tables can seat 6 people, and the rest of them can only seat 4 people. if a

maximum of 76 people can be seated at camille's cafe, how many 6-people tables does the cafe have?
Mathematics
1 answer:
morpeh [17]3 years ago
5 0
The correct anwser is 12. The solution is 76/6 then you know it's 12.666666666... But we can't say that's 13 because they are human. So wecan only say the number is 12 not 13. 
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Math number 8. <br><br> Find the value of x
ruslelena [56]

Answer:

8

Step-by-step explanation:

By intersecting chords theorem:

14(2x + 2) = 12(2x + 5) \\  \\  28x + 28 = 24x + 60 \\  \\ 28x - 24x = 60 - 28 \\  \\ 4x = 32 \\  \\ x =  \frac{32}{4}  \\  \\ x = 8

5 0
3 years ago
14.) AXYZ is shown.
Aleksandr-060686 [28]

Answer:

XY, YZ, XZ

Step-by-step explanation:

that is the order of the sides from least to greatest

6 0
3 years ago
A girl has 11 coins in dimes and quarters. Their value is $1.55. How many quarters does she have?
igomit [66]

Answer:

3

Step-by-step explanation:

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3 0
3 years ago
Lowest common factor of 72 and 108​
melamori03 [73]

Answer:

The lowest common factor of 72 and 108 is 216

5 0
3 years ago
What is the solution to the following system?
leva [86]

Answer:

(4,3,2)

Step-by-step explanation:

We can solve this via matrices, so the equations given can be written in matrix form as:

\left[\begin{array}{cccc}3&2&1&20\\1&-4&-1&-10\\2&1&2&15\end{array}\right]

Now I will shift rows to make my pivot point (top left) a 1 and so:

\left[\begin{array}{cccc}1&-4&-1&-10\\2&1&2&15\\3&2&1&20\end{array}\right]

Next I will come up with algorithms that can cancel out numbers where R1 means row 1, R2 means row 2 and R3 means row three therefore,

-2R1+R2=R2 , -3R1+R3=R3

\left[\begin{array}{cccc}1&-4&-1&-10\\0&9&4&35\\0&14&4&50\end{array}\right]

\frac{R_2}{9}=R_2


\left[\begin{array}{cccc}1&-4&-1&-10\\0&1&\frac{4}{9}&\frac{35}{9}\\0&14&4&50\end{array}\right]


4R2+R1=R1 , -14R2+R3=R3

\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&-\frac{20}{9}&-\frac{40}{9}\end{array}\right]


-\frac{9}{20}R_3=R_3

\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&1&2\end{array}\right]


-\frac{4}{9}R_3+R_2=R2 , -\frac{7}{9}R_3+R_1=R_1


\left[\begin{array}{cccc}1&0&0&4\\0&1&0&3\\0&0&1&2\end{array}\right]


Therefore the solution to the system of equations are (x,y,z) = (4,3,2)

Note: If answer choices are given, plug them in and see if you get what is "equal to".  Meaning plug in 4 for x, 3 for y and 2 for z in the first equation and you should get 20, second equation -10 and third 15.

7 0
4 years ago
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