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3 years ago

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What defines a linear pair of angles

1 answer:

Nastasia [14]3 years ago

4 0

Answer: Two adjacent, supplementary angles. Supplementary means that they make up a degree of 180 when put together, and the are right next to each other.

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Over a 24-hour period, the temperature in a town can be modeled by one period of a sinusoidal function. The temperature measures

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how many outfits of shoes does a female have for school if she has the choice of 2 pairs of shoes, 4 skirts, and 5 tops in her w

frez [133]

Answer:

if you are asking for a math equation then 2+4+5=11

Step-by-step explanation:

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3 years ago

A baseball team played 154 regular season games. The ratio of the number of games they won to the number of games they lost was

Fittoniya [83]

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88 won and 66 lost

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Won = $\frac{154}{4+3}$ ×4 = 88 games

Lost = 154 - 88 = 66 games

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3 years ago

what equations best represents the problem? You drove for 425 miles at 60 mph. How long did it take you?

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3 years ago

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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

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

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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3 years ago