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Grace [21]
3 years ago
7

Tom has 28 milk chocolates and emily has 24 dark chocolates they have to divide these chocolates into small packets that each ha

ve the same number of dark chocolates and the same number of milk chocolates
The maximum number of packets they can make is ___To make the maximum number of packets, each packet should have___ milk chocolates and ___
dark chocolates.
Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0
28+24=52
So we make a sum of what's going to equal 52 in small packets.
I found 4x13=52. 13 would be the number of packets......
BUT!! Since it MUST be equal dark and milk and we have 28 and 26,
you would have to eliminate 1 pack because all 4 would be milk chocolate.
So that is only 12 packets.
4 per pack so that would be 2 dark and 2 milk per packet.
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F(1) = 4, f(n) = f(n<br> − 1) + 11
zvonat [6]

Answer:

Step-by-step explanation:

You don't specify what you're supposed to do, so I'll make an educated guess.  

Given the sequence f(1) = 4, f(n) = f(n  − 1) + 11, find the first 5 terms:

f(1) = 4

f(2) = f(2 - 1) + 11 = f(1) + 11 = 4 + 11 = 15

f(3) = f(2) + 11 = 15 + 11 = 26

f(4) = f(3) + 11 = 26 + 11 = 37

f(5) = 37 + 11 = 48

4 0
2 years ago
Factor completely 3x − 12.<br><br> 3(x − 4)<br><br> 3(x + 4)<br><br> 3x(−12)<br><br> Prime
garri49 [273]
3x - 12
3(3x/3 - 12/3)
3(x - 4)

The answer is: 3(x - 4).
6 0
2 years ago
Read 2 more answers
( 5.6 - 1.2 ) times 1/3 =
vladimir1956 [14]

Answer:

4.4* 1/3

4.4= 4 and 4/10 so 44/10*1/3= 44/30= 22/15

22/15

Step-by-step explanation:

6 0
3 years ago
X+2y=8 x-2y=-4 solve by system of equations
Vsevolod [243]

Answer:

Step-by-step explanation:

x + 2y = 8 -----------(I)

x - 2y  = -4 -----------(II)

Add equation (I) & (II) and thus y will be eliminated and we can get the value of 'x'

(I)           x + 2y = 8

(II)          <u>x - 2y  = -4</u><u>      </u>{Now add}

           2x         = 4

                   x   = 4/2

x = 2

Substitute x =  2 in equation (I)

2 + 2y = 8

    2y = 8 - 2

    2y = 6

      y = 6/2

y = 3

3 0
2 years ago
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>

     6    -5     1       0

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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