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3 years ago

Evaluate BD for A = 5, B = -2, C = 4 and D = -6.

Mathematics

2 answers:

bezimeni [28]3 years ago

5 0

B*D?
B=-2
D=-6
B*D=(-2)*(-6)=12

krok68 [10]3 years ago

4 0

The answer is 12. Hope it helped. If you need more Details let me know. :)

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iVinArrow [24]

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valentina_108 [34]

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

$x^{3}+4^{3}\\$

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

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