Step-by-step explanation:
1. 1st of all calculate the gradient
( - 3, 5) ( 2, 10)
Gradient = (10 - 5) / ( 2--3)
= 1
2. Then find the eq
Y = mx + c
Where m is the gradient
y= 1x + c
Now replace any 2 coordinates from above in the eq.
For ex I'm taking (2, 10)
Y = 1x + c
In the coordinate, x = 2 and y =10
By replacing this in the eq, I will find c
10 = 1(2) + c
2 + c = 10
c = 10 - 2
= 8
So the eq is y = x + 8 ⬅️
Answer:
41.74 and 41.89
Step-by-step explanation:
Read https://www.wyzant.com/resources/answers/2997/how_do_you_figure_out_how_to_determine_the_best_method... and I think that will answer your question (:
Question:
A solar lease customer built up an excess of 6,500 kilowatts hour (kwh) during the summer using his solar panels. when he turned his electric heat on, the excess be used up at 50 kilowatts hours per day
.
(a) If E represents the excess left and d represent the number of days. Write an equation for E in terms of d
(b) How much of excess will be left after one month (1 month = 30 days)
Answer:
a.
b.
Step-by-step explanation:
Given
Excess = 6500kwh
Rate = 50kwh/day
Solving (a): E in terms of d
The Excess left (E) in d days is calculated using:
The expression uses minus because there's a reduction in the excess kwh on a daily basis.
Substitute values for Excess, Rate and days
Solving (b); The value of E when d = 30.
Substitute 30 for d in
<em>Hence, there are 5000kwh left after 30 days</em>