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ExtremeBDS [4]
3 years ago
6

Find the distance between the two given points H(-3,6) and C(-3,-3)

Mathematics
1 answer:
sweet [91]3 years ago
3 0

Answer:

Similarly, the distance between two points P1 = (x1,y1,z1) and P2 = (x2,y2,z2) in xyz-space is given by the following generalization of the distance formula, d(P1,P2) = (x2 x1)2 + (y2 y1)2 + (z2 z1)2. This can be proved by repeated application of the Pythagorean Theorem.

Step-by-step explanation:

Use a calculator  

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(3) Differentiating both sides of

2x^{3/2} + y^{3/2} = 29

with respect to <em>x</em> gives

3x^{1/2} + \dfrac32 y^{1/2} \dfrac{\mathrm dy}{\mathrm dx} = 0

Solve for d<em>y</em>/d<em>x</em> :

\dfrac32 y^{1/2} \dfrac{\mathrm dy}{\mathrm dx} = -3x^{1/2} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{-3x^{1/2}}{\frac32y^{1/2}} = \dfrac{-2x^{1/2}}{y^{1/2}} = -2\sqrt{\dfrac xy}

Then the slope of the tangent line to the curve at (1, 9) is

\dfrac{\mathrm dy}{\mathrm dx} = -2\sqrt{\dfrac19} = -\dfrac23

The equation of the tangent line would then be

<em>y</em> - 9 = -2/3 (<em>x</em> - 1)   ==>   <em>y</em> = -2/3 <em>x</em> + 29/3

(4) The slope of the tangent line to

y=\dfrac{ax+1}{x-2}

at a point <em>(x, y)</em> on the curve is

\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{a(x-2)-(ax+1)}{(x-2)^2} = -\dfrac{2a+1}{(x-2)^2}

When <em>x</em> = -1, we have a slope of 2/3, so

-(2<em>a</em> + 1)/(-1 - 2)² = 2/3

Solve for <em>a</em> :

-(2<em>a</em> + 1)/9 = 2/3

2<em>a</em> + 1 = -18/3 = -6

2<em>a</em> = -7

<em>a</em> = -7/2

7 0
2 years ago
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