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yuradex [85]
3 years ago
9

if maruanne father calls and says he needs 13 pounds of potting soil, how many additional bags should she buy?

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
5 0
2 bags. She needs to buy 2 bags of 10 pounds of potting soil.

20-13=7

2 more bags will give Marunne's father all the soil he needs and 7 pounds left over.
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[x^2+2y]^3<br> expand it
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I need help finding this answer because I am pretty sure it’s C but I don’t know
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Answer: A) -9

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((60 points to whoever answers all of these))
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3 years ago
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Find the roots of:
Elena-2011 [213]

\large\displaystyle\text{$\begin{gathered}\sf \pmb{1) \  2x^3-7x^2+8x-3=0 } \end{gathered}$}

Synthetic division is used since the equation is of the third degree. The divisors of -3 are 1, -1, 3, +3. So:

  | 2  -7    8  -3

<u>1 |      2   -5   3</u>

  | 2   -5    3  0

<u> 1 |      2     -3   </u>

    2   -3     0

So the factorization is (x-1)² (2x-3)=0. So:

                     \bf{ x_1=x_2=1 \qquad x_2=\dfrac{3}{2}  }

\large\displaystyle\text{$\begin{gathered}\sf \pmb{2) \  x^3-x^2-4=0 } \end{gathered}$}

Synthetic division is used since the equation is of the third degree. The divisors of -4 are 1, -1, 2, -2, 4, -4. So:

      |  1  -1  0  -4

  <u>2  |     2  2     </u>

         1  2  2  0

So the factorization is (x-2)(x²+x+2)=0 . When calculating the discriminant of the trinomial, it is concluded that it has no roots since the result is negative. So you only have one solution.

                   \bf{ 1^2-4(2)(2)=1-16=-15 < 0 \quad \Longrightarrow \quad x=2 }

\large\displaystyle\text{$\begin{gathered}\sf \pmb{3) \  6x^3+7x^2-9x+2=0 } \end{gathered}$}

Synthetic division is used since the equation is of the third degree. The divisors of 2 are 1, -1, 2, -2. So:

   | 6    7       9      2

<u>-2 |      -12    10     -2</u>

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So the factorization is (x+2)(6x²-5x+1)=0 . The quadratic equation is solved by the general formula:

         \bf{ x_{2, 3}&=\dfrac{5\pm \sqrt{(5)^2-4(6)(1)}}{2(6)}=\dfrac{5\pm \sqrt{25-24}}{12}=\dfrac{5\pm 1}{12} }}

                     \large\displaystyle\text{$\begin{gathered}\sf  \begin{matrix} x_1=-2&\ \ \ \ \ \ x_{2}=\dfrac{6}{12} \qquad &\ \ \ x_3=\dfrac{4}{12}\\ &\ \ \ x_2=\dfrac{1}{2} \qquad &x_3=\dfrac{1}{3} \end{matrix} \end{gathered}$}

6 0
2 years ago
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