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Vikki [24]
2 years ago
7

Solve for x 6x + 9 < 63

Mathematics
2 answers:
alisha [4.7K]2 years ago
8 0

6x + 9 < 63 \\ 6x < 63 - 9 \\ 6x < 54 \\ x < 9

Solution : ]-♾️ , 9[

Please give brainliest

ZanzabumX [31]2 years ago
3 0

Answer:

x < 9

Step-by-step explanation:

Given inequality:

6x + 9 < 63

Subtract 9 from both sides:

⇒ 6x + 9 - 9 < 63 - 9

⇒ 6x < 54

Divide both sides by 6:

⇒ 6x ÷ 6 < 54 ÷ 6

⇒ x < 9

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Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)
vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

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

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

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\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}

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

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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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