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aksik [14]
3 years ago
14

The value of the surface area (in square centimetres) of the cone is equal to the value of the volume (in cubic centimetres) of

the cone. The formula for the surface area S of a right cone is S=πr2+πrl, where r is the radius of the base and l is the slant height. Find the height of the cone.
Height:
cm

Please help Thank you

Mathematics
1 answer:
aliina [53]3 years ago
3 0

Answer: The height of the cone is 10.9 cm

Step-by-step explanation:

Step 1: Find the height of the cone by Applying the Pythagoras Theorem

12² = 5² + h²

144 = 25 + h²

h² = 144 - 25

h² = 119

h = 10.9 cm

So the height of the cone is 10.9 cm

Hope this helps!

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What is the area of the sector with a central angle of 49° and a radius of 11 cm? Use 3.14 for pi and round your final answer to
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\begin{array}{rcl}A& = &3.14\times 11^{2}\times\dfrac{49}{360}\\\\ & = &3.14 \times 121 \times 0.1361\\ & = & \mathbf{51.71}\\\end{array}\\\text{The area of the sector is }\boxed{\textbf{51.71 cm}^{2}}

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\large\begin{array}{l}\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}\,dx}\\\\ =\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{-1}{\sqrt{1-x^2}}\cdot (-2x)\,dx\qquad\quad(i)} \end{array}


\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{1-x^2=u~~\Rightarrow~~-2x\,dx=du}\\\\\\ \textsf{Finding the new limits of integration:}\\\\ \begin{array}{lcl} \textsf{When }\mathsf{x=0}&~\Rightarrow~&\mathsf{u=1-0^2}\\\\ &&\mathsf{u=1}\\\\\\ \textsf{When }\mathsf{x=\dfrac{1}{2}}&~\Rightarrow~&\mathsf{u=1-\left(\dfrac{1}{2}\right)^2}\\\\ &&\mathsf{u=1-\dfrac{1}{4}}\\\\ &&\mathsf{u=\dfrac{3}{4}} \end{array} \end{array}


\large\begin{array}{l} \textsf{Then (i) becomes}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}}\frac{-1}{\sqrt{u}}\,du}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}}\frac{-1}{u^{\frac{1}{2}}}\,du}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}} (-1)\cdot u^{-\frac{1}{2}}\,du}\\\\ =\mathsf{(-1)\cdot \dfrac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg|_{1}^{\frac{3}{4}}} \end{array}

\large\begin{array}{l} =\mathsf{(-1)\cdot \dfrac{~~u^{\frac{1}{2}}~~}{\frac{1}{2}}\bigg|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{(-1)\cdot 2u^{\frac{1}{2}}\Big|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{-2\cdot \sqrt{u}\Big|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{-2\cdot \left(\sqrt{\dfrac{3}{4}}-\sqrt{1}\right)} \end{array}

\large\begin{array}{l} =\mathsf{-2\cdot \left(\dfrac{\sqrt{3}}{2}-1\right)}\\\\ =\mathsf{-\diagup\!\!\!\! 2\cdot \dfrac{\sqrt{3}}{\diagup\!\!\!\! 2}+(-2)\cdot (-1)}\\\\ =\mathsf{-\sqrt{3}+2}\\\\ =\mathsf{2-\sqrt{3}} \end{array}


\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}\,dx=2-\sqrt{3}} \end{array}}\qquad\checkmark \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2153950


\large\textsf{I hope it helps.}


</span><span><span>Tags: <em>definite integral integrate limits function irrational square root sqrt fraction composite substitution integral calculus</em></span>
</span>
4 0
4 years ago
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