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satela [25.4K]
2 years ago
13

Find the slope for (3,1)and (4,3)

Mathematics
2 answers:
nadezda [96]2 years ago
8 0

Answer:

2

Step-by-step explanation:

slope formula : \frac{y_2-y_1}{x_2-x_1} where the x and y values are derived from known points

We are given the points (3,1)and (4,3)

Given these points we can define our variables

(x1,y1) = (3,1) so x1 = 3 and y1 = 1

(x2,y2) = (4,3) so x2 = 4 and y2 = 3

Now that we have defined our variable, we plug them into the formula

Recall formula: \frac{y_2-y_1}{x_2-x_1}

x1 = 3 , y1 = 1 , x2 = 4 , y2 = 3

* plug in values *

\frac{3-1}{4-3}

subtract 3 and 1 , and 4 and 3

\frac{3-1=2}{4-3=1} =\frac{2}{1} =2

the slope is 2

Vaselesa [24]2 years ago
3 0
Y2-y1 / x2-x1

so, 3-1 =2 and 4-3=1
2/1 is 2
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777dan777 [17]

Answer: t1 = -34; t2 = -64; t3 = -94

d is the distance between numbers (d > 0)

we have:

t5 = t1 + 4d = -154

t9 = t1 + 8d = -274

we have the equations:

t1 + 4d = -154

t1 + 8d = -274

<=> t1 = -34

      d = -30

with t1 = -34 and = -30 => t2 = -34 - 30 = -64

                                          t3 = -64 - 30 = -94

Step-by-step explanation:

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3 years ago
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
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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

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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

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x² = 81

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

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

                                    = 237.09 cm²

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Step-by-step explanation:

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