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vampirchik [111]
3 years ago
15

A telephone company charges a fixed monthly rate plus a rate per minute of usage. The company charges $155 for 100 minutes of us

age and $245 for 200 minutes of usage. An equation can be written to show the relationship between the total minutes used (x) and the total monthly charges (y). Which of the following best describes the steps to draw the graph of y against x?
A) draw a graph which joins the points (155, 100) and (245, 200) and has a slope = 0.9
B) draw a graph which joins the points (155, 100) and (245, 200) and has a slope = 1.1
C) draw a graph which joins the points (100, 155) and (200, 245) and has a slope = 1.1
D) draw a graph which joins the points (100, 155) and (200, 245) and has a slope = 0.9
Mathematics
2 answers:
torisob [31]3 years ago
8 0

Answer:

The answer is D.

Step-by-step explanation:

As said in the problem, we are labeling the total minutes used <em>x </em>and the total monthly income as <em>y</em>. This means that in the graph we will be making the y-axis should represent the total monthly charges, and the x-axis should represent the total minutes.

Thus the graph should go through points (100,155) and (200,245) and its slope is:

\frac{245-155}{200-100} =0.9.

Vinil7 [7]3 years ago
3 0

Answer:

The answer would be D according the the question at hand.

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Simplify : 48÷6 [-52 ÷2 {4 - 3 (2 - 15÷3)}]​
garri49 [273]

Answer:

86

Step-by-step explanation:

48÷6 [-52 ÷2 {4 - 3 (2 - 15÷3)}] \\ 8[-52 ÷2 {4 - 3 (2 - 15÷3)}] \\ 8[-52 ÷2{1 (2 - 15÷3)}] \\ 8[ - 2.5{ (2 - 15÷3)}] \\ 8[ - 2.5{ ( - 13÷3)}] \\ 8[ - 2.5{ (  - 4.3)}] \\ 8[10.75 { }] \\  = 86

8 0
2 years ago
Find the solution set of this inequality|10x+20| ≤10
Pavlova-9 [17]

Answer:

solution is

[-3,-1]

Step-by-step explanation:

we are given

|10x+20|\leq 10

Firstly, we will find critical values

so, let's assume it is equal

|10x+20|= 10

now, we can break absolute sign

For |10x+20|= -(10x+20):

-(10x+20)= 10

we can solve for x

-10x-20= 10

Add both sides by 20

-10x-20+20= 10+20

-10x= 30

Divide both sides by -10

and we get

x=-3

For |10x+20|= (10x+20):

(10x+20)= 10

we can solve for x

10x+20= 10

Subtract both sides by 20

10x+20-20= 10-20

10x= -10

Divide both sides by 10

and we get

x=-1

so, critical values are

x=-3

x=-1

now, we can draw a number line and locate these values

and then we can check inequality on each intervals

For (-\infty,-3):

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-5

|10\times -5+20|\leq 10

|-50+20|\leq 10

30\leq 10

so, this is FALSE

For [-3,-1]:

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-2

|10\times -2+20|\leq 10

|-20+20|\leq 10

0\leq 10

so, this is TRUE

For (-1,\infty):

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=0

|10\times 0+20|\leq 10

|0+20|\leq 10

20\leq 10

so, this is FALSE

so, solution is

[-3,-1]

7 0
3 years ago
You are charged 6.5% tax on a 42% purchase. Find the amount of tax
Rus_ich [418]
Assuming you means 6.5% tax of a $42 purchase

0.065*42=2.73

tax is $2.73
7 0
3 years ago
Please put a serious answer
aksik [14]

Answer:

x=12

Step-by-step explanation:

9x-14=2(3x+11)

9x-14=6x+22

9x-6x=22+14

3x=36

x=36/3=12

8 0
3 years ago
A box's volume is equal to 600 cubic
olga2289 [7]

Answer:

Step-by-step explanation:

Length = x

width = x - 7

height = x - 10

volume = 600 in^3

Volume = L * w * h

600 = x * (x - 7)*(x - 10)

600 = x * (x^2 - 17x + 70)

600 = x^3 - 17x^2 + 70x

x^3 - 17x^2 + 70x - 600 = 0

There is a genius somewhere that could solve this using algebra. The rest of us have 2 choices. We can graph it (see below) or we could use a program that solves it for us. I'll do both for you.

a = 1

b  = - 17

c = 70

d = - 600

There is only 1 real solution: x = 15

L = 15

w = 8

h = 5

That has to be one of the ugliest graphs made. It does into the hundreds of thousands and it still looks like a straight line. Trust me. It's not.

3 0
2 years ago
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