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mojhsa [17]
3 years ago
8

Help I will mark brain list!!!!!

Mathematics
1 answer:
dangina [55]3 years ago
3 0

Answer: I'm not sure exactly how you want this answered.

Step-by-step explanation:

Just by simplifying the expression, I got \frac{a^{2}xy }{2} +a^{3}x-\frac{3x^{2} y}{5} -ay^{2}

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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

7 0
3 years ago
Write an equation in standard form of a quadratic function that has x-intercepts -2 and 5 and k=-4
bulgar [2K]

Answer:

-2x + 5 = -4k

Step-by-step explanation:

8 0
3 years ago
Solve.<br> 92.5÷105 brainliest
Karo-lina-s [1.5K]

Answer:

\frac{37}{42}

Step-by-step explanation:

Given :

\frac{92.5}{105}

Multiply numerator and denominator by 2 :

\frac{92.5}{105} \times\frac{2}{2}

\frac{185}{210}          [Divide both sides by 5 as it is a common factor]

\frac{185}{210} \div \frac{5}{5}

\frac{37}{42}

6 0
2 years ago
Read 2 more answers
The graphs below show four functions: Graph A shows function f of x equals 4 multiplied by 3 to the power of x. Graph B shows fu
s2008m [1.1K]

Answer:

Graph B shows function f of x equals 4 multiplied by 3 to the power of negative x.

Step-by-step explanation:

f(x) = 4(3)^{-x} is the value 4 times three raised to the negative x. Graph B is the best representation.

5 0
3 years ago
Read 2 more answers
A newspaper’s cover page is 3/8 text, and photographs fill the rest. If 2/5 of the text is an article about endangered species,
jolli1 [7]

Answer:

\frac{3}{20}

Step-by-step explanation:

Let x be the cover page

We are given that a newspaper 's cover page is \frac{3}{8} text and photograph fill the rest.

We have to find that how much part of cover's page is the article about endangered species

Text =\frac{3}{8}x

Remaining part =x-\frac{3}{8} x

Remaining part =\frac{8x-3x}{8}=\frac{5}{8} x

Therefore, a newspaper;s cover page fill with  Photograph =\frac{5}{8}

If the article about endangered species=\frac{2}{5} of text

Therefore, the article about endangered species=\frac{2}{5}\times\frac{3x}{8}=\frac{3}{20}x

Hence,  the article about endangered species is \frac{3}{20}of the cover page.

7 0
3 years ago
Read 2 more answers
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