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GalinKa [24]
2 years ago
8

Find the values of x1 and x2 where the following two constraints intersect.

Mathematics
1 answer:
Arte-miy333 [17]2 years ago
4 0

Answer: x1 = 251/26, x2 = -111/26

Step-by-step explanation:

Hi!

As you can see in the figure, the point you are looking for is the intersection of two lines.

The intersection point is found solving this system of linear equations (the point must satisfy both equations):

9x_1 +7x_2=57\\4x_1 + 6x_2 = 13

You can solve it, for example, by the method of substitution:

\text{solve for x1 in the first equation:}\\x_1 = \frac{1}{9}(57 - 7x_2)

Then plug x1 into equation 2, and solve for x2:

\frac{4}{9}(57-7x_2) + 6x_2 = 13\\\text{doing the algebra you get:}\\x_2 = \frac{-111}{26}

Then you use the value of x2 to get x1:

x_1 = \frac{1}{9}(57 - 7x_2)= \frac{1}{9}(57 + 7*\frac{111}{26}) = 251/26\\

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                   Dimensions of square has length of 10 cm

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Step:2

                  Formula for area of square,

                                                 A_{square} = a^{2}

                 where, a - side of length = 10 cm

                                                             =10^{2}

                                                            =100

                    Area of square, A_{square} = 100 Square centimeter

Step:3

                Formula for area of Trapezoid,

                                            A_{Trapezoid} = \frac{a+b}{2} h.......................................(2)

                    Where, a - 8 cm

                                 b - 14 cm

                                 h - 4 cm

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                                                            = (11)(4)

                                                            = 44

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             Formula for area of rectangle,

                                            A_{Rectangle} =lb......................................(3)

            Where, l = 20 cm

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           Equation (3) becomes,

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