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NARA [144]
3 years ago
11

Consider the curve x+xy+2y^2=6. the slope of the line tangent to the curve at the point (2,1) is?

Mathematics
2 answers:
oksian1 [2.3K]3 years ago
7 0
<span>The derivative of your function is:  
1 + (1*dy/dx + y) - 4y(dy/dx) = 0
2 + dy/dx + y - 4y(dy/dx) = 0
(2 + y) + dy/dx (1 - 4y) = 0
(dy/dx) = -(2 + y)/(1 - 4y)
 
and at point (2,1),
dy/dx = -(2 + 1)/(1 - 4*1)
          = -3 / -3
          = 1
 
Slope of tangent line = 1</span>
viktelen [127]3 years ago
5 0
The derivative is actually (-y-1)/(4y+x)
then you plug in the points (2,1) ==>
(-1-1)/(4*1+2) which equals -1/3
then you just plug in your info into a point-slope formula and get:
y-1=-1/3(x-2) which can then be reduced
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The opposite of squaring a number is it square root it.

Step-by-step explanation:

7 0
3 years ago
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Suppose $p(x)$ is a monic cubic polynomial with real coefficients such that $p(3-2i)=0$ and $p(0)=-52$. determine $p(x)$ (in exp
Maslowich

If p(3-2i)=0, then complex number 3-2i is a root of cubic polynomial.

If polynomial has real coefficients, then conjugated 3+2i is also a root of polynomial.

Then the polynomial will be of a form

p(x)=(x-(3-2i))(x-(3+2i))(x-a).

Since p(0)=-52, then

-52=(0-(3-2i))(0-(3+2i))(0-a),\\ \\-52=(2i-3)(3+2i)a,\\ \\-52=(4i^2-9)a,\\ \\-52=(-4-9)a,\\ \\-13a=-52,\\ \\a=4.

Therefore,

p(x)=(x-(3-2i))(x-(3+2i))(x-4),\\ \\p(x)=(x^2-6x+13)(x-4),\\ \\p(x)=x^3-10x^2+37x-52.

3 0
3 years ago
What's the answer ? To this question
Airida [17]
You could use the equation y=mx+b

Find the slope of the two coordinates. I would personally suggest using stack and subtract:

(0,2) 2-5 -3 3
(2,5) -------= ------= -----
0-2 -2 2

We have the slope but we need to find the y-intercept.

y=3
--x+b
2

Plug one of the coordinates in the constructed equation above:

2=3
-- (0)+b
2

2=0+b
-0=-0
------------
b=2

We found the slope and y-intercept. Now we have our complete equation. The equation is:

y=3
--x+2
2


Have a blessed day!
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Bye measuring by a ruler or something that is equivalent to 8 inches<span />
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