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DanielleElmas [232]
3 years ago
9

16x+3-14x+15 simplify this expression

Mathematics
2 answers:
MissTica3 years ago
4 0

Answer:

<<<<<<It is 2x+18>>>>>>


Ilya [14]3 years ago
3 0

Answer:

\boxed{=2x+18}

Step-by-step explanation:

You can used their group like terms.

16x-14x+3+15

Then you add by the similar elements.

16-14=2

2x+3+15

Finally you add by the numbers.

15+3=18

=2x+18

Hope this helps!

And thank you for posting your question at here on brainly, and have a great day.

-Charlie

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Help please! I’ll give brainliest.
Lapatulllka [165]

9514 1404 393

Answer:

  (a) not proportional

  (b) k = 7/8

Step-by-step explanation:

When a linear equation is of the form ...

  y = mx + b

with b ≠ 0, the relation is NOT PROPORTIONAL.

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If b=0, so the equation is of the form ...

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then the relation IS PROPORTIONAL. The constant of proportionality is k.

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(a) Not proportional

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8 0
3 years ago
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

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.

8 0
3 years ago
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