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MAXImum [283]
3 years ago
14

You evenly split the bill 135 among 3 people you all agree to leave a 20% tip of your bill. How much does each person owe toward

s the bill and tip
Mathematics
1 answer:
lana66690 [7]3 years ago
6 0
135/3=45
20% of 45=9
45+9=54
so each person owes $54 toward bill and tip
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The solutions for each case are listed below:

  1. x = 65
  2. x = 35
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<h3>How to solve on systems of linear equation by taking advantage of angle relationships</h3>

In this problem we must solve algebraic equations by taking advantage of angle properties. Now we proceed to solve the variables for each case:

Case 1 - Opposite angles

2 · x - 10 = 120

2 · x = 130

x = 65

Case 2 - Opposite angles

2 · x + 25 = 3 · x - 10

25 + 10 = 3 · x - 2 · x

35 = x

x = 35

Case 3 - Opposite angles generated by two perpendicular lines

2 · y + 50 = x + 44                      (1)

5 · y - 17 = 7 · x - 248                 (2)

- x + 2 · y = - 6

7 · x - 5 · y = 231

x = 48, y = 21

Case 4 - Opposited angles generated by two perpendicular angles

6 · x = 90                     (3)

9 · y + 18 = 90              (4)

The solution to this system of linear equations is (x, y) = (15, 8).

To learn more on systems of linear equations: brainly.com/question/21292369

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2 years ago
An isosceles triangle has base angles measuring 25 degrees each. What is the measure of the vertex angle?
Yuri [45]
A triangle's sum must add to 180 degrees. If two of the angles, summed, equal 50 degrees, then the vertex angle must equal 130 degrees. So, the vertex angle equals 130 degrees.
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Kindly, check the attachment below to see the answer.

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Find a1, for the arithmetic series<br> with S14 = - 420 and d = -6.
faust18 [17]

\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} n = 14\\ d= -6 \end{cases} \\\\\\ a_{14}=a_1+(14-1)(-6)\implies a_{14}=a_1+(13)(-6)\implies a_{14}=a_1-78 \\\\[-0.35em] ~\dotfill

\bf \textit{sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ \cline{1-1} n= 14\\ S_{14}=-420\\ a_{14}=a_1-78 \end{cases}\implies S_{14}=\cfrac{n(a_1+a_{14})}{2} \\\\\\ -420=\cfrac{14[a_1+(a_1-78)]}{2}\implies -420=7(2a_1-78)\implies \cfrac{-420}{7}=2a_1-78 \\\\\\ -60=2a_1-78\implies 18=2a_1\implies \cfrac{18}{2}=a_1\implies 9=a_1

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