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zhenek [66]
1 year ago
8

You started this year with $458 saved and

Mathematics
2 answers:
vovangra [49]1 year ago
6 0

Answer:

s = 14m + 458

Step-by-step explanation:

current amount = 458

14 per month

s = 14m + 458

Alex787 [66]1 year ago
6 0
The equation has to be put into a Y=mx+b format. Basically warp this equation to fit your needs. Make the Y (y intercept) equal S. m (slope) equal $14. x (x coordinate value) equal months or m. Then the b equal $458.

So essentially it will be S=$14m+$458.

It’s this because S is the total value you’re trying to get (the savings)

$14 is linear meaning it’s the same every time or in this case every month so it goes there since the amount also depends on the amount of months. Which is also why m (months) is being multiplied by $14. $458 goes there because it’s the starting value so it will be accounted for every time you calculate with this equation.

I apologize if anything is incorrect or unclear and if you have any questions please ask.
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Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).
guapka [62]

Answer:

(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)

We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

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... 64 = y³ . . . . . . multiply by 4y, add 64

... 4 = y . . . . . . . . cube root

... y²/4 = 16/4 = x = 4

_____

2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

Multiplying by 8 gives ...

... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

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