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fenix001 [56]
3 years ago
14

Write an equation of the line below

Mathematics
1 answer:
gulaghasi [49]3 years ago
5 0

Answer:

y = 4/5x - 2

Step-by-step explanation:

To create an equation with just a line compare two points on the graph and see how far the x and y points are on the graph then put them into a fraction with y over x then find the y intercept or find where the x coordinate equals 0 in this case it was (0, -2). Now put it all together:

y = mx + b

mx = comparison of two points

b = y intercept

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Maria has a cell phone and she gets charged for data each month. She is
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If it costs $12.5 for every GB and we use 8 then 12.5*8=$100. So if our flat rate now is $70 we would loose 100-70=$30

So Maria would loose $30 a month if she pays per GB
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3 years ago
the cost of having one shirt laundered is 6 dollars. each additional shirt costs 4 dollars. write the explicit and recursive for
Katarina [22]

Answer:

<h3>explicit formula- a(n)=6+4(n-1)</h3><h3>recursive formula- a(n)=a(n)-1 +4; a(1)=6</h3>

Step-by-step explanation:

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3 years ago
Hie, here's my question: $1800 at 6.5% for 30 months
miv72 [106K]

Answer:

A. 10105.86 (compound) or 3510 (simple)

B.11905.86 (compound) or 5310 (simple)

Step-by-step explanation:

If its's compound interest:

1800(1 +  \frac{6.5}{100}) ^{30} = 11905.86

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7 0
3 years ago
A new building that costs $1,000,000 has a useful life of 10 years and a scrap value of $200,000. Using straight-line depreciati
vazorg [7]

Answer:

V = -80,000*t + 1,000,000

Step-by-step explanation:

Given:

- The original cost V_i = $1,000,000

- Scrap Value V_f = $200,000

- Total number of years n = 10

Find:

Using straight-line depreciation, find the equation for the value V in terms of t, where t is in years.

Solution:

- The straight line depreciation method means their is a linear relationship between the value of building and time elapsed till useful life. The linear relationship takes a form:

                                 V = m*t + C

Where,

V: It is the current value of the building

m: The rate at which value increase or decrease with time t

C: The initial value of the building

- We will compute constants m and C as follows:

                               m = (V_f - V_i) / ( 10 - 0 )

                               m = (200,000 - 1,000,000) / 10

                               m = - $80,000/year

                               C = V_i = $1,000,000

- Plug the values of the constants evaluated above in the linear expression:

                                V = -80,000*t + 1,000,000

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