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Murljashka [212]
3 years ago
5

A parallelogram is drawn on a

Mathematics
1 answer:
mezya [45]3 years ago
6 0

Answer:

(-3 , -3)

Step-by-step explanation:

A: (-1,2), B: (1,2), and C: (-1,-3)  .... D: (x,y)

slope of AB = (2-2)/(1 - -1) = 0    .... parallel to x axis

x coordinate of A is 2 units left to B (-1 vs 1)

y coordinate of C is 2 units left to B so D must be 2 units left to C: -1 -2 = <u>-3</u>

DC // AB , y coordinate of D = C =<u> -3</u>

D: (-3 , -3)

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Find the equation of the tangent of x^2- xy + y^2 =7 at (-1, 2)
Citrus2011 [14]

Answer:

<u>It</u><u> </u><u>is</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>4</u><u>x</u><u> </u><u>+</u><u> </u><u>1</u><u>0</u>

Step-by-step explanation:

Let's first get the slope of the curve.

[ slope is the derivative of the equation ]

{x}^{2}  - xy +  {y}^{2}  = 7

introduce dy/dx :

\frac{d}{dx} ( {x}^{2}  - xy +  {y}^{2} ) =  \frac{d}{dx} (7) \\  \\ 2x - (y +  \frac{dy}{dx} ) + 2y \frac{dy}{dx}  = 0

make dy/dx the subject:

2x - y -  \frac{dy}{dx}  + 2y \frac{dy}{dx}  = 0 \\  \\ 2y \frac{dy}{dx}  -  \frac{dy}{dx}  = y - 2x \\  \\  \frac{dy}{dx} (2y - 1) = y - 2x \\  \\  \frac{dy}{dx}  =  \frac{y - 2x}{2y - 1}

At point (-1, 2):

\frac{dy}{dx}  =  \frac{2 - 2( - 1)}{2(2) - 1}  \\  \\ slope =  \frac{4}{3}

but a tangent has the same slope as the curve:

y = mx + c

m is the slope

c is the y-intercept

At (-1, 2):

2 = ( - 1 \times  \frac{4}{3} ) + c \\  \\ c = 2 +  \frac{4}{3}  \\  \\ c =  \frac{10}{3}

equation:

y =  \frac{4}{3} x +  \frac{10}{3}  \\  \\ { \boxed{3y = 4x + 10}}

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8.5 * 14.5 = 123.25 dollars

D.

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2 years ago
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