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3 years ago

Solve for z. -cz+6z= tz+83

2 answers:

5 0

-cz + 6z = tz + 83.....subtract tz from both sides
-cz + 6z - tz = 83....factor out z on the left side
z(-c + 6 - t) = 83.....divide both sides by (-c + 6 - t)
z = 83 / (-c + 6 - t)

Likurg_2 [28]3 years ago

3 0

Hello i saw you question and I knew the answer was this, it is a fraction

z= v-83
-----
c+t-6

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Korolek [52]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Sum of cubes: (a b)(a2 – ab b2) = a3 b3 Difference of cubes: (a – b)(a2 ab b2) = a3 – b3 Which products result in a sum or diffe

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

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

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