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3 years ago

A rectangular shaped probe part is 10 cm long, 5 cm wide, and 5 cm high. What is the volume of this part?

Mathematics

1 answer:

lapo4ka [179]3 years ago

3 0

Answer:

V= L x W x H

L=10cm

W=5cm

H=5cm

·.·V=10cm x 5cm x 5cm

Step-by-step explanation:

The volume of a regular object is the product of the object's length , width(breadth) and height.

= L x W x H

and since the length is 10cm ,the height 5cm and width 5cm,multiple the three.

=10cm x 5cm x 5cm

The product is 250 because 10 x 5 is 50 and then 50 is multiplied by 5 which is 250

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By comparing all the options with the formula of sum of cubes and difference of cubes we get that $(x+4)\left(x^{2}-4 x+16\right)$ is sum of cubes

<h3>What is formula of sum of cubes and difference of cubes?</h3>

Formula of sum of cubes and difference of cubes are following

$(a+b)(a^2 -ab +b^2) = a^3+ b^3 \\\\\\(a - b)(a^2+ ab+ b^2) = a^3 - b^3$

(a)

$(x-4) \left(x^{2}+4 x-16\right)\\=(x-4) \left(x^{2}+4 x-4^2\right)\\\\$

by comparison with formula we can say that this is neither sum or nor difference of cubes.

(b)

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

by comparison with formula we can say that this is neither sum nor difference of cubes.

(c)

$(x+1)(x-1)=x^{2}-1^{2}\\$

we can say that this is neither sum nor difference of cube as it's difference of squares.

(d)

$(x+4)\left(x^{2}-4 x+16\right)\\\\(x+4)\left(x^{2}-4 x+4^2\right)$

by comparison with formula this is equal to

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Hence this is sum of cubes

(e)

$(x+4)\left(x^{2}+4 x+16\right)\\(x+4)\left(x^{2}+4 x+4^2\right)$

by comparison with formula we can say that this is neither sum nor difference of cubes.

By comparing all the options with the formula of sum of cubes and difference of cubes we get that $(x+4)\left(x^{2}-4 x+16\right)$ is sum of cubes

To know more about cubes visit: brainly.com/question/107100

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2 years ago

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Answer:

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