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Artyom0805 [142]
3 years ago
7

Write an equation for your trend line?

Mathematics
1 answer:
Soloha48 [4]3 years ago
4 0
Find two points on the graph that the line crosses through almost perfectly. It looks like (1,10) and (9,1) will do.

Use them to compute the slope:
m = (1 - 10) / (9 - 1)
= -9/8

Then set up the "point-slope form":
y - y0 = m * (x - x0)

You choose some point (x0, y0) that the line crosses through. We already know the line passes through (1,10) pretty well, so let's use that.

x0 = 1
y0 = 10

Now finish plugging into the equation:
y - 10 = -9/8 * (x - 1)

The above equation will work fine for an answer, but let's go a step further and solve for y.

y - 10 = -9/8x + 9/8
y = -9/8x + 9/8 + 10
y = -9/8x + 9/8 + 80/8
y = -9/8x + (9 + 80)/8
y = -9/8x + 89/8
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Consider the expression 625(5xy)^-3/ (5x)^2 y^7
andre [41]
<span>625(5xy)^-3/ (5x)^2 y^7
     
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= --------------------   /  25x^2y^7
      125 x^3y^3

= 5/x^3y^3    / </span>25x^2y^7
= 5/x^3y^3   *  (1/ 25x^2y^7)
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answer


         1
-----------------
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