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uranmaximum [27]
3 years ago
7

Jackson buys tickets for two concerts the tickets for the concert featuring the local artist for $25 each the tickets for the co

ncert featuring the international artists are $75. Jackson by the total of 20 tickets. How many tickets does Jackson by for the concert featuring international artist?
Mathematics
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

I believe he bought  18

Step-by-step explanation:

if he bought 20 in total and you remove the two from local artists

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What value of x is in the solution set of 3(x – 4) ≥ 5x + 2? <br><br> –10<br> –5<br> 5<br> 10
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<u>Answer:</u>

The value of x is in the solution set of 3(x – 4) ≥ 5x + 2 is -10

<u>Solution:</u>

Need to determine which value of x from given option is solution set of 3(x – 4) ≥ 5x + 2

Lets first solve 3(x – 4) ≥ 5x + 2

3(x – 4) ≥ 5x + 2

=> 3x – 12 ≥ 5x + 2

=> 3x – 5x ≥ 12 + 2

=> -2x ≥ 14

=> -x ≥ 7

=> x ≤ -7

All the values of x which are less than or equal to -7 is solution set of 3(x – 4) ≥ 5x + 2. From given option there is only one value that is -10 which is less than -7  

Hence from given option -10 is solution set of 3(x – 4) ≥ 5x + 2.

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\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-4}{6-1}\implies \cfrac{-5}{5}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=-1(x-1) \\\\\\ y-4=-x+1\implies y=-x+5


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\bf (\stackrel{x_1}{-12}~,~\stackrel{y_1}{14})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-14}{6-(-12)}\implies \cfrac{-15}{6+12}\implies \cfrac{-15}{18}\implies -\cfrac{5}{6}


\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-14=-\cfrac{5}{6}[x-(-12)] \\\\\\ y-14=-\cfrac{5}{6}(x+12)\implies y-14=-\cfrac{5}{6}x-10\implies y=-\cfrac{5}{6}x+4

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