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Dima020 [189]
2 years ago
10

The London theater offers child tickets at 10$each and adult tickets at 20$ each.Mrs burham buys 34 tickets at a total cost of 5

70 if C is the number of child tickets and A is the number of adult tickets write the system of equations which represents this situation
Mathematics
1 answer:
kirill115 [55]2 years ago
4 0

Answer:

A+C=34

10C+20A=570

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Pani-rosa [81]

Answer:

its the second one.

Step-by-step explanation:

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3 years ago
What is the horizontal asympote of this graph?
Anna35 [415]

Using it's concept, it is found that the graph has no horizontal asymptote.

<h3>What are the horizontal asymptotes of a function f(x)?</h3>

The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity.

In this problem, we have that:

  • The function is undefined for x < 0, hence \lim_{x \rightarrow -\infty} f(x) is undefined.
  • For x > 0, the funciton goes to infinity, hence \lim_{x \rightarrow \infty} f(x) = \infty.

Thus, the graph has no horizontal asymptote.

More can be learned about horizontal asymptotes at brainly.com/question/16948935

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2 years ago
Put in order from least to greatest 12, 8, -9, -12, 10, 16 *​
Nina [5.8K]

Answer:

-12,-9,8,10,12,16

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
CON PROCESOS POR FAVOR
labwork [276]

Answer:

\dfrac{149}{40}

\dfrac{38}{15}

Step-by-step explanation:

(\dfrac{7}{8} + \dfrac{4}{5}) - (\dfrac{9}{20} + \dfrac{-5}{2}) =

= (\dfrac{7}{8} \times \dfrac{5}{5} + \dfrac{4}{5} \times \dfrac{8}{8}) - (\dfrac{9}{20} + \dfrac{-5}{2} \times \dfrac{10}{10})

= (\dfrac{35}{40} + \dfrac{32}{40}) - (\dfrac{9}{20} + \dfrac{-50}{20})

= \dfrac{67}{40} - (-\dfrac{41}{20})

= \dfrac{67}{40} + \dfrac{41}{20} \times \dfrac{2}{2}

= \dfrac{67}{40} + \dfrac{82}{40}

= \dfrac{149}{40}

(-\dfrac{6}{4} + \dfrac{3}{2}) + (\dfrac{6}{5} + \dfrac{4}{3}) =

= (-\dfrac{3}{2} + \dfrac{3}{2})+ (\dfrac{6}{5} \times \dfrac{3}{3} + \dfrac{4}{3} \times \dfrac{5}{5})

= 0 + \dfrac{18}{15} + \dfrac{20}{15}

= \dfrac{38}{15}

4 0
2 years ago
Combine the following fractions by addition or subtraction as directed.
Marina86 [1]
Answer:\frac{6}{y^2-xy}-\frac{6}{x^2-xy}=\frac{6(x+y)}{xy(y-x)}

Explanation:

Combining the fractions, we have:

\begin{gathered} \frac{6(x^2-xy)-6(y^2-xy)}{(x^2-xy)(y^2-xy)} \\  \\ =\frac{6x^2-6xy-6y^2+6xy}{(x^2-xy)(y^2-xy)} \\  \\ =\frac{6x^2-6y^2}{x(x-y).y(y-x)} \\  \\ =\frac{6(x-y)(x+y)}{-xy(x-y)^2} \\  \\ =\frac{-6(x+y)}{xy(x-y)} \\  \\ =\frac{6(x+y)}{xy(y-x)} \end{gathered}

8 0
1 year ago
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