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Tresset [83]
3 years ago
6

Tres por un número que es agregado a otro numero es 18. El doble del primer número menos el otro numero es 12. Encuentra los num

ero
Mathematics
1 answer:
andre [41]3 years ago
5 0
Creo que la respusta es 6 porque 3 multiplicado por 6 es 18 y 6 multiplicado por 2 es 12.18 menos 12 es 6.
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If 80% of a number is 20, what is 20% of the number?
Vlad1618 [11]

Answer: 5

Step-by-step explanation:

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25/5=5

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3 years ago
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timurjin [86]
4.5 ... 9/2 and you will get the answer
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Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold: $\bullet$ $f(x)$ is divisible by $x^3$. $\bullet
frosja888 [35]

There seems to be one character missing. But I gather that <em>f(x)</em> needs to satisfy

• x^3 divides f(x)

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f(x) = x^5 + \cdots

Since x^3 divides f(x), and

f(x) = x^3 p(x)

where p(x) is degree-2, and we can write it as

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Now, we have

f(x)^2 = \left(x^3p(x)\right)^2 = x^6 p(x)^2

so if (x - 1)^3 divides f(x)^2, then p(x) is degree-2, so p(x)^2 is degree-4, and we can write

p(x)^2 = (x-1)^3 q(x)

where q(x) is degree-1.

Expanding the left side gives

p(x)^2 = x^4 + 2ax^3 + (a^2+2b)x^2 + 2abx + b^2

and dividing by (x-1)^3 leaves no remainder. If we actually compute the quotient, we wind up with

\dfrac{p(x)^2}{(x-1)^3} = \underbrace{x + 2a + 3}_{q(x)} + \dfrac{(a^2+6a+2b+6)x^2 + (2ab-6a-8)x +2a+b^2+3}{(x-1)^3}

If the remainder is supposed to be zero, then

\begin{cases}a^2+6a+2b+6 = 0 \\ 2ab-6a-8 = 0 \\ 2a+b^2+3 = 0\end{cases}

Adding these equations together and grouping terms, we get

(a^2+2ab+b^2) + (2a+2b) + (6-8+3) = 0 \\\\ (a+b)^2 + 2(a+b) + 1 = 0 \\\\ (a+b+1)^2 = 0 \implies a+b = -1

Then b=-1-a, and you can solve for <em>a</em> and <em>b</em> by substituting this into any of the three equations above. For instance,

2a+(-1-a)^2 + 3 = 0 \\\\ a^2 + 4a + 4 = 0 \\\\ (a+2)^2 = 0 \implies a=-2 \implies b=1

So, we end up with

p(x) = x^2 - 2x + 1 \\\\ \implies f(x) = x^3 (x^2 - 2x + 1) = \boxed{x^5-2x^4+x^3}

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Gemiola [76]
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skelet666 [1.2K]

Answer:1

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