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3 years ago

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The Area Ratio of two similar solids is 169: 289. If the volume of the smaller solid is 689,858 cm, what is the volume of the la

rger solid?

1 answer:

masha68 [24]3 years ago

5 0

Answer:

1,542,682 cm^3.

Step-by-step explanation:

The scale factor = √169:√289

= 13:17

So the Volume ratio is 13^3:17^3

So the volume of the larger solid is 689,858 * (17^3/13^3)

= 1,542,682 cm^3.

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ValentinkaMS [17]

Answer:

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Step-by-step explanation:

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(x^2 - x^(1/2))/(1-x^(1/2))

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$\frac { \left( { x }^{ 2 }-{ x }^{ \frac { 1 }{ 2 } } \right) }{ \left( 1-{ x }^{ \frac { 1 }{ 2 } } \right) }$

$\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x } \right) }{ \left( 1-\sqrt { x } \right) } \cdot 1$

$\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x } \right) }{ \left( 1-\sqrt { x } \right) } \cdot \frac { \left( 1+\sqrt { x } \right) }{ \left( 1+\sqrt { x } \right) }$

$\\ \\ =\frac { { x }^{ 2 }+{ x }^{ 2 }\sqrt { x } -\sqrt { x } -x }{ 1+\sqrt { x } -\sqrt { x } -x }$

$\\ \\ =\frac { -\sqrt { x } \left( 1-{ x }^{ 2 } \right) -x\left( 1-x \right) }{ \left( 1-x \right) }$

$\\ \\ =\frac { -\sqrt { x } \left( 1+x \right) \left( 1-x \right) -x\left( 1-x \right) }{ \left( 1-x \right) }$

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3 years ago

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Vinil7 [7]

7. X2=25
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3 years ago

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What’s this answer ????

slega [8]

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3 years ago

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

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3 years ago