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san4es73 [151]
2 years ago
6

The sum of three numbers is -39.22. One of the numbers is 22.8. The other two numbers are equal to each other. What is the value

of the other two numbers?
Mathematics
1 answer:
masya89 [10]2 years ago
7 0

Answer:

The value of the other two numbers is 31.01

Step-by-step explanation:

Let the two numbers equal to each other be x

One of the numbers is 22.8

The sum of three numbers is -39.22.

So, x+x+22.8=-39.22

2x+22.8=-39.22

2x=-39.22-22.8

2x=-62.02

x=\frac{-62.02}{2}

x=-31.01

Hence the value of the other two numbers is 31.01

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Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

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3 years ago
You run 4and3/4 miles through a park each day if you ran that distance five times last week how many miles did you run
zzz [600]
23.75 miles, because you would take 4 and 3/4 (4.75) and multiply by 5 :)
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3x(x²-5)

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Answer:-2/3

Step-by-step explanation:

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