Find the slope of the line that will go through point (4, 5) and (8, 9) first and then you can use the point-slope form to figure out the equation of the line. Since I am not sure what form of the equation they want, I will just provide you the point-slope form, standard form, and the slope-intercept form of the line.
The slope can be calculated by subtracting the difference of the y-coordinates of the two points and then dividing it by the difference of the x-coordinates.
(y_2 - y_1) / (x_2 - x_1) = slope
(9 - 5) / (8 - 4) = 4 / 4 = 1
The point-slope form is: y - y_1 = m(x - x_1)
Let's take the coordinate (4, 5) and use 4 for x_1 and 5 for y_1.
y - 5 = 1(x - 4)
The slope-intercept form is: y = mx + b
We can modify the point-slope form to look like the slope-intercept form.
y - 5 = 1(x - 4)
y = x - 4 + 5
y = x + 1
The standard form is: ax + by = c
We can modify the slope-intercept form to look like the standard form.
y = x + 1
y - x = 1
-x + y = 1
Point-slope form: y - 5 = 1(x - 4)
Slope-intercept form: y = x + 1
Standard form: -x + y = 1
I believe its A please tell me if I'm wrong its been a while
Will start to distribute the 2 in (x-7)
so it can be (2x-2*7)=100
(2x-14)=100
2x=100+14
2x=114
x=114/2
x=57
The initial investment = $250
<span>annual simple interest rate of 3% = 0.03
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Let the number of years = n
the annual increase = 0.03 * 250
At the beginning of year 1 ⇒ n = 1 ⇒⇒⇒ A(1) = 250 + 0 * 250 * 0.03 = 250
At the beginning of year 2 ⇒ n = 2 ⇒⇒⇒ A(2) = 250 + 1 * 250 * 0.03
At the beginning of year 3 ⇒ n = 3 ⇒⇒⇒ A(2) = 250 + 2 * 250 * 0.03
and so on .......
∴ <span>The formula that can be used to find the account’s balance at the beginning of year n is:
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A(n) = 250 + (n-1)(0.03 • 250)
<span>At the beginning of year 14 ⇒ n = 14 ⇒ substitute with n at A(n)</span>
∴ A(14) = 250 + (14-1)(0.03*250) = 347.5
So, the correct option is <span>D.A(n) = 250 + (n – 1)(0.03 • 250); $347.50
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