64. Susan tipped at the rate of 75 dollars to 5 waiters.
Question: She can afford to pay up to 90 dollars, How many waiters can she tipped if that is the case.
Let’s solve first and identify how much will she be giving to each waitress with 75 dollars is to 5 waiters rate:
=> 75 dollars / 5 waiters = 15 dollars each waiter
Now, she have 90 dollars
=> 90 – 75 = 15
Thus
=> 90 / 15 = 6 waiters she will be tipping.
Answer:
1.
5
x
−
2
y
=
4
; (−1, 1)
2.
3
x
−
4
y
=
10
; (2, −1)
3.
−
3
x
+
y
=
−
6
; (4, 6)
4.
−
8
x
−
y
=
24
; (−2, −3)
5.
−
x
+
y
=
−
7
; (5, −2)
6.
9
x
−
3
y
=
6
; (0, −2)
7.
1
2
x
+
1
3
y
=
−
1
6
; (1, −2)
8.
3
4
x
−
1
2
y
=
−
1
; (2, 1)
9.
4
x
−
3
y
=
1
;
(
1
2
,
1
3
)
10.
−
10
x
+
2
y
=
−
9
5
;
(
1
5
,
1
10
)
11.
y
=
1
3
x
+
3
; (6, 3)
12.
y
=
−
4
x
+
1
; (−2, 9)
13.
y
=
2
3
x
−
3
; (0, −3)
14.
y
=
−
5
8
x
+
1
; (8, −5)
15.
y
=
−
1
2
x
+
3
4
;
(
−
1
2
,
1
)
16.
y
=
−
1
3
x
−
1
2
;
(
1
2
,
−
2
3
)
17.
y
=
2
; (−3, 2)
18.
y
=
4
; (4, −4)
19.
x
=
3
; (3, −3)
20.
x
=
0
; (1, 0)
Find the ordered pair solutions given the set of x-values.
21.
y
=
−
2
x
+
4
; {−2, 0, 2}
22.
y
=
1
2
x
−
3
; {−4, 0, 4}
23.
y
=
−
3
4
x
+
1
2
; {−2, 0, 2}
24.
y
=
−
3
x
+
1
; {−1/2, 0, 1/2}
25.
y
=
−
4
; {−3, 0, 3}
26.
y
=
1
2
x
+
3
4
; {−1/4, 0, 1/4}
27.
2
x
−
3
y
=
1
; {0, 1, 2}
28.
3
x
−
5
y
=
−
15
; {−5, 0, 5}
29.
–
x
+
y
=
3
; {−5, −1, 0}
30.
1
2
x
−
1
3
y
=
−
4
; {−4, −2, 0}
31.
3
5
x
+
1
10
y
=
2
; {−15, −10, −5}
32.
x
−
y
=
0
; {10, 20, 30}
Find the ordered pair solutions, given the set of y-values.
33.
y
=
1
2
x
−
1
; {−5, 0, 5}
34.
y
=
−
3
4
x
+
2
; {0, 2, 4}
35.
3
x
−
2
y
=
6
; {−3, −1, 0}
36.
−
x
+
3
y
=
4
; {−4, −2, 0}
37.
1
3
x
−
1
2
y
=
−
4
; {−1, 0, 1}
38.
3
5
x
+
1
10
y
=
2
; {−20, −10, −5}
Part B: Graphing Lines
Given the set of x-values {−2, −1, 0, 1, 2}, find the corresponding y-values and graph them.
39.
y
=
x
+
1
40.
y
=
−
x
+
1
41.
y
=
2
x
−
1
42.
y
=
−
3
x
+
2
43.
y
=
5
x
−
10
44.
5
x
+
y
=
15
45.
3
x
−
y
=
9
46.
6
x
−
3
y
=
9
47.
y
=
−
5
48.
y
=
3
Find at least five ordered pair solutions and graph.
49.
y
=
2
x
−
1
50.
y
=
−
5
x
+
3
51.
y
=
−
4
x
+
2
52.
y
=
10
x
−
20
53.
y
=
−
1
2
x
+
2
54.
y
=
1
3
x
−
1
55.
y
=
2
3
x
−
6
56.
y
=
−
2
3
x
+
2
57.
y
=
x
58.
y
=
−
x
59.
−
2
x
+
5
y
=
−
15
60.
x
+
5
y
=
5
61.
6
x
−
y
=
2
62.
4
x
+
y
=
12
63.
−
x
+
5
y
=
0
64.
x
+
2
y
=
0
65.
1
10
x
−
y
=
3
66.
3
2
x
+
5
y
=
30
Part C: Horizontal and Vertical Lines
Find at least five ordered pair solutions and graph them.
67.
y
=
4
68.
y
=
−
10
69.
x
=
4
70.
x
=
−
1
71.
y
=
0
72.
x
=
0
73.
y
=
3
4
74.
x
=
−
5
4
75. Graph the lines
y
=
−
4
and
x
=
2
on the same set of axes. Where do they intersect?
76. Graph the lines
y
=
5
and
x
=
−
5
on the same set of axes. Where do they intersect?
77. What is the equation that describes the x-axis?
78. What is the equation that describes the y-axis?
Part D: Mixed Practice
Graph by plotting points.
79.
y
=
−
3
5
x
+
6
80.
y
=
3
5
x
−
3
81.
y
=
−
3
82.
x
=
−
5
83.
3
x
−
2
y
=
6
84.
−
2
x
+
3
y
=
−
12
Step-by-step explanation:

Total amount of Brian paid is $48.30.
Hope this helps. - M
The answer is 1.
Because if you simplify (divide) 19/19 you get one.
4/4 would also be 1. Any number divided to itself will always be 1.
Whereas 30/15 would be 2 because thirty divided by fifteen is 2
Answer:
it would help her know how to prepare her teaching to match the students learning and expectations
Step-by-step explanation:
This idea of opening this tutoring service for students in these grades would prove a success if if martine has adequate knowledge of her students/customers. That is the learners requirements, their expectations, their experiences, and their strengths and weaknesses in particular subject areas.
Knowledge of these expectations would help to set Martine on the path of tutoring success and this would attract more students. So for her to have a strong tutoring business she has to know the approaches to use to make students strong academically, and how to match learning ability with her teaching.