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Dimas [21]
2 years ago
13

Help please!! Thank you!! :)

Mathematics
1 answer:
Naddik [55]2 years ago
5 0

Answer:

  A.  m^2+7m+10=0

Step-by-step explanation:

This is a problem in pattern matching, and in substituting a variable for a pattern.

  (x^2+3)^2 +7x^2 +21 = -10 . . . . . . given

  (x^2 +3)^2 +7(x^2 +3) = -10 . . . . . factor the last two terms

  m^2 +7m = -10 . . . . . . . . . . . . subsitute m for x^2 +3

  m^2 +7m +10 = 0 . . . . . . . . add 10 to both sides; matches A

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Answer:

The answer is D. 15.65.

Step-by-step explanation:

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Find an equation of the tangent line to the curve 2(x^2+y^2)2=25(x^2−y^2) (a lemniscate) at the point (3,1)
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\bf 2[x^2+y^2]^2=25(x^2-y^2)\qquad \qquad 
\begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
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\bf \left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[x-y\frac{dy}{dx}  \right]
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notice... a derivative is just the function for the slope

now, you're given the point 3,1, namely x = 3 and y = 1

to find the "m" or slope, use that derivative, namely f'(3,1)=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}

that'd give you a value for the slope

to get the tangent line at that point, simply plug in the provided values
in the point-slope form

\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad
\begin{cases}
x_1=3\\
y_1=1\\
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\end{cases}\\ \qquad \uparrow\\
\textit{point-slope form}

and then you solve it for "y", I gather you don't have to, but that'd be the equation of the tangent line at 3,1

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