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Doss [256]
3 years ago
13

Solve x2 + 5x = −6. (6 points)

Mathematics
2 answers:
Damm [24]3 years ago
6 0

Answer:

<em><u>x = -0.5</u></em><em> </em>

Step-by-step explanation:

<em>You need to combine like terms</em>

x * 2 + 5x = -6

<em>Add x to 5x</em>

6x * 2 = -6

<em>Divide 2 from both sides</em>

6x = -3

<em>Divide 6 from both sides to get x alone</em>

x = -0.5

<em>Hope this helps</em>

  • (\) QueTooOfficial (/)
Mumz [18]3 years ago
3 0

Answer:

S = {-2, -3}

Step-by-step explanation:

x² + 5x = -6 is the start of a quadratic equation

to solve: bring -6 to the left side by adding 6, so the right side is equal to 0

x² + 5x + 6 = 0

there are multiple ways to solve this equation, two common ways you could use to solve is the quadratic formula or factoring. in this equation, factoring works and is a much shorter process than using the quadratic formula.

to factor is to basically break down the equation. so to break down the equation, we need to find what 2 numbers multiplied together equal 6, and what numbers added together equal 5? both numbers have to be the same numbers added together and multiplied. as in, you cant use 4 + 1 to make 5 and 2 times 3 to make 6.

the answer to this is 2 and 3, as 2 + 3 = 5 and 2 x 3 = 6.

so the equation replaced by factoring would be (x + 2)(x + 3) = 0.

where do the x's come from?: they come from x², one <em>x</em> for each factor.

to check if this factoring job works, you can use the FOIL method to check, which multiples the First two terms, the two Outside terms, the two Inside terms, and the Last two terms of each factor.

moving on to the next step in finding the final answer, you would set (x + 2) and (x + 3) equal to 0 and solve for x

x + 2 = 0 <-- subtract 2 from both sides to get x alone

x = -2

x + 3 = 0 <-- subtract 3 from both sides to get x alone

x = -3

so the solution to this equation would be {-2,-3}

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Answer:

(d)  \displaystyle 12x^3 - 15x^2 + 2

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<u>Algebra I</u>

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<u>Calculus</u>

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Derivative Notation

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Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

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