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 #1
avatar+9491 
+2

area of lower base  =  225π cm2

                                                                               and...

area of lower base  =  π(radius of lower base)2

                                                                               So...

π(radius of lower base)2  =  225π cm2

                                                                    Divide both sides of the equation by  π .

(radius of lower base)2  =  225 cm2

                                                                    Take the positive square root of both sides.

radius of lower base  =  15 cm

 

 

area of upper base  =  25π cm2

                                                                               and...

area of upper base  =  π(radius of upper base)2

                                                                               So...

π(radius of upper base)2  =  25π cm2

                                                                    Divide both sides by  π .

(radius of upper base)2  =  25 cm2

                                                                    Take the positive square root of both sides.

radius of upper base  =  5 cm

 

 

So this is what a cross section of the cone perpendicular to the base looks like:

 

 

where  h  is the altitude of the small cone that was cut off, and each length is in cm.

 

The bases of the frustrum are parallel, and corresponding angles are congruent,

So by AA similarity, we can say that the large right triangle and the small right triangle are similar.

So...

 

\(\frac{\text{height of large triangle}}{\text{height of small triangle}}=\frac{\text{base of large triangle}}{\text{base of small triangle}}\\~\\ \frac{24+h}{h}=\frac{15}{5}\\~\\ \frac{24+h}{h}=3\\~\\ 24+h=3h\\~\\ 24=2h\\~\\ 12=h\)

.
Jun 19, 2018
Jun 18, 2018
 #3
avatar
+1

Thanks :D

Jun 18, 2018
 #1
avatar+26400 
+4

10 sin^2(x) + 10 sin (x) cos (x) - cos^2 (x) = 2.

Solve for all values of  x between 0 and 360 degrees

 

\(\begin{array}{|rcll|} \hline 10 \sin^2(x) + 10 \sin (x) \cos (x) - \cos^2 (x) &=& 2 \\ && \small{\boxed{\sin(x) = \tan(x)\cos(x)}} \\ 10 [\tan(x)\cos(x)]^2 + 10 \tan(x)\cos(x)\cos (x) - \cos^2 (x) &=& 2 \\ 10 \tan^2(x)\cos^2(x) + 10 \tan(x)\cos^2(x) - \cos^2 (x) &=& 2 \\ \cos^2(x) \Big( 10 \tan^2(x) + 10 \tan(x) - 1 \Big) &=& 2 \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 \cdot \dfrac{1}{\cos^2(x) } \\ && \small{\boxed{ \dfrac{1}{\cos^2(x) } = 1+\tan^2(x)} } \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 \cdot( 1+\tan^2(x)) \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 +2\tan^2(x) \\ 8 \tan^2(x) + 10 \tan(x) - 3 &=& 0 \\ && \large{\boxed{\tan(x) = z}} \\ 8 z^2 + 10z - 3 &=& 0 \\\\ z&=& \frac{-10\pm \sqrt{100-4\cdot 8\cdot(-3)} } {2\cdot 8} \\ z&=& \frac{-10\pm \sqrt{196} } {16} \\ z&=& \frac{-10\pm 14 } {16} \\\\ z_1 &=& \dfrac{-10 + 14 } {16} \\ \mathbf{z_1} &\mathbf{=}& \mathbf{\dfrac{1} {4}} \\\\ z_2 &=& \dfrac{-10 - 14 } {16} \\ \mathbf{z_2} &\mathbf{=}& -\mathbf{\dfrac{3} {2}} \\ \hline \end{array}\)

 

solutions:

\(\begin{array}{|rcll|} \hline \tan(x) &=& z_1 \\ \tan(x) &=& \frac{1}{4} \\ \mathbf{x} &\mathbf{=}& \mathbf{\arctan(\frac{1}{4}) + n\cdot 180^{\circ} } \quad & | \quad n \in Z \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \tan(x) &=& z_2 \\ \tan(x) &=& -\frac{3}{2} \\ x &=& \arctan(-\frac{3}{2}) + n\cdot 180^{\circ} \\ \mathbf{x} &\mathbf{=}& \mathbf{-\arctan(\frac{3}{2}) + n\cdot 180^{\circ} } \quad & | \quad n \in Z \\ \hline \end{array}\)

 

 x between 0 and \( \mathbf{360^{\circ}}\)

\(\begin{array}{|rcll|} \hline x_1 &=& \arctan(\frac{1}{4}) \\ \mathbf{x_1} &\mathbf{=}& \mathbf{14.0362434679^{\circ}} \\\\ x_2 &=& -\arctan(\frac{3}{2}) + 180^{\circ} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{123.690067526^{\circ}} \\\\ x_3 &=& \arctan(\frac{1}{4})+ 180^{\circ} \\ \mathbf{x_3} &\mathbf{=}& \mathbf{194.036243468^{\circ}} \\\\ x_4 &=& -\arctan(\frac{3}{2}) + 2\cdot 180^{\circ} \\ \mathbf{x_4} &\mathbf{=}& \mathbf{303.690067526^{\circ}} \\ \hline \end{array}\)

 

 

laugh

Jun 18, 2018
 #4
avatar+118724 
0

Thanks :)

Jun 18, 2018
 #1
avatar+9491 
+2

A point of intersection is a point that makes both equations true.

 

y  =  3x2 + 4x - 5       and       y  =  x2 + 11

 

Starting with the first equation...
 

y  =  3x2 + 4x - 5

                                       Since  y  =  x2 + 11  we can substitute  x2 + 11  in for  y .

x2 + 11  =  3x2 + 4x - 5

                                       Subtract  x2  from both sides of the equation.

11  =  2x2 + 4x - 5

                                       Subtract  11  from both sides of the equation.

0  =  2x2 + 4x - 16

                                       Divide through by  2 .

0  =  x2 + 2x - 8

                                       Factor the right side.

0  =  (x - 2)(x + 4)

                                       Set each factor equal to zero and solve for  x .

 

x - 2  =  0          or          x + 4  =  0

 x  =  2              or             x  =  -4

 

Use these values of  x  to find  y .

 

If    x  =  2    then    y  =  (2)2 + 11  =  4 + 11  =  15    So   (2, 15)  is a point of intersection.

 

If    x  =  -4    then    y  =  (-4)2 + 11  =  16 + 11  =  27    So   (-4, 27)  is a point of intersection.

 

The points of intersection are   (-4, 27)   and   (2, 15) .

Jun 18, 2018
Jun 17, 2018

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