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Vesna [10]
4 years ago
8

Helen has 4 jogging outfits and three pairs of shoes. How many different outfits can she make?

Mathematics
2 answers:
Afina-wow [57]4 years ago
4 0
She can make 12 different outfits

alexgriva [62]4 years ago
4 0
She can wear 12 different outfits. 
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Race to get brainliest, need help!
Gemiola [76]

Answer:

Original number is 86

Step-by-step explanation:

              Original:   10x + y

Double reversed:   2(10y + x)

Constraints:

  x + y = 14                                     ===>            x = 14 - y

  2(10y + x) + 10x + y = 222          ===>           21y + 12x = 222

Substitution to find y:

  21y + 12x = 222

  21y + 12(14 - y) = 222

  21y + 168 - 12y = 222

            9y + 168 = 222

                      9y = 54

                        y = 6

Substitution to find x:

  x = 14 - y

  x = 14 - 6

  x = 8

Original:   10x + y

                 10(8) + (6)

                 80 + 6

                 86

7 0
3 years ago
Read 2 more answers
What is the value of x?<br> to<br> x = [?]°<br> o
Ugo [173]

Answer:

The value of x from the figure is 58 degrees

To get the value of x, we will use the formula:

The angle at the vertex = 1/2(difference of angles at the arc)

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51 = 1/2(160 - x)

102 = 160 - x

x = 160 - 102

x = 58 degrees

Hence the value of x from the figure is 58 degrees

6 0
3 years ago
A side of a square is 10 m longer than the side of an equilateral triangle the perimeter of the square is three times the perime
Slav-nsk [51]

Answer:

\large \boxed{\text{8 m}}

Step-by-step explanation:

     Let side of triangle = x

And side of rectangle = y

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We have two conditions:

\begin{array}{lrcl}(1) & y & = & x + 10\\(2) &4y & = & 3\times 3x\\& 4y & = & 9x\\& 4(x + 10) & = & 9x\\& 4x + 40 & = & 9x\\& 40 & = & 5x\\& x& = &\mathbf{8}\\\end{array}\\\text{Each side of the triangle is $\large \boxed{\textbf{8 m}}$ long.}

3 0
3 years ago
What is the answer to 7/9+(-5/12)
galina1969 [7]
13 / 36 :)




bxksms
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8 0
3 years ago
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solniwko [45]

Step-by-step explanation:

please complete your question

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