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AveGali [126]
3 years ago
7

Use points (2001,17.60) and (2002,18.75) to write an equation in slope intercept form.

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
3 0
The general slope-intercept form is given by y = mx + b, where m is the slope and b is the y-intercept.

Let's find the slope first. Remember that the slope given by two points is \frac{y_2 - y_1}{x_2 - x_1}. So, using the two given points,

m = \frac{y_2 - y_1}{x_2 - x_1}
m = \frac{18.75 - 17.60}{2002 - 2001}
m = 1.15

Now, we know that y = 1.15 x + b. To find b, let's just plug in the x and y coordinates for one of the points and solve for b:

17.60 = 1.15(2001) + b
17.60 = 2301.15 + b
b = -2283.55

Thus, the equation of the line in slope-intercept form is

\bf y = 1.15x - 2283.55
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The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct
lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

y=\dfrac{x}{1+x}

Point P\left(1,\dfrac{1}{2}\right) lies on the curve.

Point Q\left(x,\dfrac{x}{1+x}\right) is an arbitrary point on the curve.

The slope of the secant line PQ is

\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]

1. If x=0.5, then the slope is

\dfrac{1}{2(0.5+1)}=\dfrac{1}{3}\approx 0.3333

2. If x=0.9, then the slope is

\dfrac{1}{2(0.9+1)}=\dfrac{1}{3.8}\approx 0.2632

3. If x=0.99, then the slope is

\dfrac{1}{2(0.99+1)}=\dfrac{1}{3.98}\approx 0.2513

4. If x=0.999, then the slope is

\dfrac{1}{2(0.999+1)}=\dfrac{1}{3.998}\approx 0.2501

5. If x=1.5, then the slope is

\dfrac{1}{2(1.5+1)}=\dfrac{1}{5}\approx 0.2

6. If x=1.1, then the slope is

\dfrac{1}{2(1.1+1)}=\dfrac{1}{4.2}\approx 0.2381

7. If x=1.01, then the slope is

\dfrac{1}{2(1.01+1)}=\dfrac{1}{4.02}\approx 0.2488

8. If x=1.001, then the slope is

\dfrac{1}{2(1.001+1)}=\dfrac{1}{4.002}\approx 0.2499

7 0
3 years ago
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