Can someone help me on this question

6, 10, 14, ...<br> Find the 30th term.

finlep [7]

50 i think that’s it sorry if it’s wrong

8 0

3 years ago

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How are the properties of segments and angles used to determine their measure?

zavuch27 [327]

You have to be able to know the appropriate properties or postulates to apply on the angles or segments to simplify them. An example would be the Angle Addition Postulate. This postulates states that the sum of the individual measures of the interior angles is equal to the measure of the included angle. There are still a lot more properties to choose from depending on the situation.

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E5_1%20%7B%20%5Cfrac%7BlnR%7D%7B%20R%5E%7B2%7D%20%20%7D%20%5C%2C%20dR%20"

alex41 [277]

Integrate by parts, setting

$\begin{matrix}u=\ln R&&\mathrm dv=\dfrac{\mathrm dR}{R^2}\\\mathrm du=\dfrac{\mathrm dR}R&&v=-\dfrac1R\end{matrix}$

So the integral is

$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\dfrac{\ln R}R\bigg|_{R=1}^{R=5}+\int_1^5\frac{\mathrm dR}{R^2}$
$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\left(\frac{\ln5}5-\frac{\ln1}1)-\frac1R\bigg|_{R=1}^{R=5}$
$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\frac{\ln5}5-\left(\frac15-1\right)$
$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=\frac15(4-\ln5)$

5 0

4 years ago

The manager of a store wants to have a sales promotion where she gives prizes to the first several people through the door. She

Viktor [21]

The answer is 60 people. Each person would get <span>4 pins, 6 ornaments and 9 mugs.</span>

4 0

4 years ago

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Find an equation of the tangent line to the hyperbola x2 a2 − y2 b2 = 1 at the point (x0, y0).

nalin [4]

The equation of the tangent line is $y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

To find the tangent to the hyperbola $\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }$ at (x₀, y₀), we differentiate the equation implicitly to find the equation of the tangent at (x₀, y₀).

So, $\frac{d}{dx} (\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }) = \frac{d0}{dx}\\\frac{d}{dx} \frac{x^{2} }{a^{2} } - \frac{d}{dx}\frac{y^{2} }{b^{2} }= 0\\\frac{2x }{a^{2} } - \frac{dy}{dx}\frac{2y }{b^{2} } = 0\\\frac{2x }{a^{2} } = \frac{dy}{dx}\frac{2y }{b^{2} } \\\frac{dy}{dx} = \frac{b^{2}x }{a^{2}y }$

So, at (x₀, y₀)

$\frac{dy}{dx} = \frac{b^{2} x_{0} }{a^{2} y_{0} }$

So, the equation of the tangent line is gotten from the standard equation of a line in point-slope form

So, $\frac{y - y_{0} }{x - x_{0} } = \frac{b^{2} x_{0} }{a^{2} y_{0} } \\y - y_{0} = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) \\y = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) + y_{0} \\y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

So, the equation of the tangent line is $y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

Learn more about equation of tangent line here:

brainly.com/question/12561220

3 0

3 years ago