Match each equation to its factorized version and solution.

Mathematics

1 answer:

bagirrra123 [75]3 years ago

5 0

Answer:

$24x -{6x}^{2} = 0 \: matches \: with \: 6x(4 - x) \: and \: the \: solution \: x = 0 \ \: or \: \: x = 4$

$2 {x}^{2} + 6x = 0 \: matches \: with \: 2x(x + 3) = 0 \: and \: the \: solution \: x = 0 \: or \: x = - 3$

$4x - {x}^{2} = 0 \: matches \: with \: x(4 - x) = 0 \: and \: the \: solution \: x = 0 \: or \: x = 4$

$14x - {7x}^{2} = 0 \: matches \: with \: 7x(2 - x) = 0 \: and \: the \: solution \: x = 0 \: or \: x = 2$

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Veseljchak [2.6K]

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4 years ago

24 , Factor: x²-9 A (x + 3XX-3) B (x+3)2 с (x-3) (x+3)(x+3)

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Answer:

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Step-by-step explanation:

We are given the expression $\displaystyle \large{x^2-9}$ :—

To factor this expression, we have a formula for it which is difference of two squares:—

$\displaystyle \large{a^2-b^2=(a+b)(a-b)}$

You can also swap from $\displaystyle \large{(a+b)(a-b)}$ to $\displaystyle \large{(a-b)(a+b)}$ via multiplication property.

From the expression, factor using the formula above:—

$\displaystyle \large{x^2-9=(x^2)-(3)^2}\\\displaystyle \large{x^2-9=(x-3)(x+3)}$

Therefore, the factored expression is:—

$\displaystyle \large{\boxed{(x-3)(x+3)}}$

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If you have any questions regarding the problem or need clarification of my answer/explanation, do not hesitate to ask in comment!

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2 years ago

How to derive: f(w)=ln(cos(w-1))

postnew [5]

The word you're looking for is "differentiate", not "derive".

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