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scZoUnD [109]
3 years ago
10

Solve n³ + 2n² - 15n = 0 by factoring. Show the factored form of the equation, and the resulting solutions.

Mathematics
2 answers:
Semenov [28]3 years ago
5 0

Answer: n = 0

n = 3

n = - 5

Step-by-step explanation:

The given cubic equation is expressed as

n³ + 2n² - 15n = 0

Since n is common to each term, we would factorize n out. It becomes

n(n² + 2n - 15) = 0

n = 0 or

n² + 2n - 15 = 0

To further factorize the quadratic equation, we would would find two numbers such that their sum or difference is 2n and their product is - 15n². The two numbers are 5n and - 3n. Therefore,

n² + 5n - 3n - 15 = 0

n(n + 5) - 3(n + 5) = 0

(n - 3)(n + 5) = 0

n = 3 or n = - 5

laila [671]3 years ago
4 0

Step-by-step explanation:

Take n common

n(n²+2n-15)=0

now factorise the quadratic one as n when goes to RHS it becomes zero

n²-3n+5n-15=0

n(n-3)+5(n-3)=0

n=3,-5

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If a line has a slope of 1, its function can be write as x-y=c

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Answer:

\begin{bmatrix}-2\\0 \end{bmatrix}

Step-by-step explanation:

We have the product, 2\times \begin{bmatrix}-1\\0 \end{bmatrix}.

It is known that,

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So, we get,

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use the picture to write each ratio as a simplified fraction (how are you going to find the length of the third side?) show your
Butoxors [25]

Answer:

<h3>See below</h3>

Step-by-step explanation:

to figure out the ratios we must figure out the length of <u>hypotenuse</u> first to do so we can consider <u>Pythagoras</u><u> theorem</u> given by

\displaystyle  {a}^{2}  +   {b}^{2}  =  {c}^{2}

\displaystyle \implies c =  \sqrt{ {a}^{2} +  {b}^{2}  }

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\displaystyle c =  \sqrt{ {12}^{2} +  {9}^{2}  }

simplify squares:

\displaystyle c =  \sqrt{ 225 }

simplify square root:

\displaystyle c = 15

now recall that,

  • \displaystyle  \sin( \theta)  =  \frac{opp}{hypo}
  • \displaystyle\cos( \theta)  =  \frac{adj}{hypo}
  • \displaystyle   \tan( \theta)  =  \frac{opp}{adj}

the ratios with respect to angle w given by

  • \displaystyle  \sin( W)  =  \frac{12}{15}  =  \frac{4}{5}
  • \displaystyle  \cos(W)  =  \frac{9}{15}  =  \frac{3}{5}
  • \displaystyle  \tan( W)  =  \frac{12}{9}  =  \frac{4}{3}

the following ratio with respect to angle X

  • \displaystyle  \sin(X)  =  \frac{9}{15}  =  \frac{3}{5}
  • \displaystyle  \cos(X)  =  \frac{12}{15}  =  \frac{4}{5}
  • \displaystyle  \tan( X)  =  \frac{9}{12}  =  \frac{3}{4}
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