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Bas_tet [7]
2 years ago
14

The surface areas of two similar solids are 169 m2 and 81 m2. the volume of the larger solid is 124.92 m3. which proportion corr

ectly shows how to solve for the volume of the smaller solid, x? = = = =
Mathematics
2 answers:
ikadub [295]2 years ago
6 0

By finding the scale factor, we will see that the volume of the smaller solid is 86.75 m³.

<h3>How to get the volume of the smaller solid?</h3>

If the solids are similar, then there is a scale factor between the two. Then the relation between the areas is equal to the scale factor squared, and the relation between the volumes is equal to the scale factor cubed.

This means that if the areas are 169 m² and 81 m², then we can write:

169 m² = (k²)*81 m²

Solving for k, we get:

k = √(169 m²/81 m²) = 1.44

Then if the volume of the large solid is 124.92m³ we can write:

124.92m³ = k³*V

Replacing k and solving for V we get:

124.92m³ = (1.44)³*V

(124.92m³/ (1.44)³) = V = 86.75 m³

If you want to learn more about scale factors:

brainly.com/question/3457976

#SPJ4

Mandarinka [93]2 years ago
3 0

Answer: Its C

Step-by-step explanation: Just did it

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Fourth box: 167 * 10^-29 meters.

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3 years ago
If FC IS RS 4430 and the BEP in unitsis 89, then contribution margin is ?
Umnica [9.8K]

Answer: RS 49.8

Step-by-step explanation:

Given the following :

Fixed cost = RS 4430

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Break even point in units = Fixed cost ÷ contribution margin

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3 years ago
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I need help with q5
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Percent increase = ((new number - original number) / (original number)) x 100.

percent increase = ((3000 - 2500) / 2500)) x 100 =
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7 0
3 years ago
Guys! Help me find how much of this circle is shaded!
Alona [7]

Answer:

The answer is 1/10

Step-by-step explanation:

Let us take the Shaded region be x

So,

2/5 + x = 1/2

Now, Solve for x

x + 2/5 = 1/2

5x + 2/5 = 1/2

2(5x + 2) = 5(1)

10x + 4 = 5

10x + 4 – 4 = 5 – 4

10x= 1

10x/10 = 1/10

x = 1/10

Thus, The value of x is 1/10

<h3><u>For</u><u> Verification</u>;</h3>

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4 + 1/10 = 1/2

5/10 = 1/2

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L.H.S = R.H.S

Hence Verified!

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