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Vinil7 [7]
3 years ago
14

In standard form: (x+5)(3x-1)-(x-4)^2

Mathematics
2 answers:
denpristay [2]3 years ago
6 0
I hope this helps you



x.3x+x.(-1)+5.3x+5.(-1)-(x^2-2.4.x+16)


3x^2-x+15x-5-x^2+8x-16


2x^2+22x-21
Minchanka [31]3 years ago
5 0
Basically, expand
expand them seperately to reduce confusion
exponent first
(x-4)^2=(x-4)(x-4)=x²-4x-4x+16=x²-8x+16

(x+5)(3x-1)=3x²+15x-x-5=3x²+14x-5

3x²+14x-5-(x²-8x+16)
3x²+14x-5-x²+8x-16
3x²-x²+14x+8x-5-16
2x²+22x-21

answer is
2x²+22x-21
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Answer:

-20a - 65 = -104  for  a = 2.7

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-20a - 50  for  a = 2.7

Insert 2.7 in for a.

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Answer with Step-by-step explanation:

We are given that an equation of curve

x^{\frac{2}{3}}+y^{\frac{2}{3}}=4

We have to find the equation of tangent line to the given curve at point (-3\sqrt3,1)

By using implicit differentiation, differentiate w.r.t x

\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0

Using formula :\frac{dx^n}{dx}=nx^{n-1}

\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}

\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}

\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}

Substitute the value x=-3\sqrt3,y=1

Then, we get

\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}

\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}

Slope of tangent=m=\frac{1}{\sqrt3}

Equation of tangent line with slope m and passing through the point (x_1,y_1) is given by

y-y_1=m(x-x_1)

Substitute the values then we get

The equation of tangent line is given by

y-1=\frac{1}{\sqrt3}(x+3\sqrt3)

y-1=\frac{x}{\sqrt3}+3

y=\frac{x}{\sqrt3}+3+1

y=\frac{x}{\sqrt3}+4

This is required equation of tangent line to the given curve at given point.

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