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IRISSAK [1]
3 years ago
11

Send help!!!!!!!!!!!!!!

Mathematics
2 answers:
Veronika [31]3 years ago
7 0
Hello !

(17^{ \frac{1}{7}})^{7}  = \\  \\   17^{ \frac{1}{7}*7 } =   \\  \\ = 17^{1} \\  \\  = 17


From the formula : (a^{m})^{n} =a^{m*n}

I hope it helped:)

Diano4ka-milaya [45]3 years ago
5 0
Powers of power multiply:

(17^{1/7})^7 = 17^{(\frac 1 7 \cdot 7)} = 17^1 = 17

Answer: 17


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Which is a perfect square?<br>61<br>62<br>63<br>65​
Morgarella [4.7K]

Answer:

None

Step-by-step explanation:

None of these, but \sqrt{64}=8 so if 64 is an option than that's the correct option

\sqrt{61} =7.81 \\\sqrt{62}=7.87\\\sqrt{63} =7.93\\\sqrt{65} =8.06

3 0
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
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
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Answer:

D) 55/73

Step-by-step explanation:

The question is on trigonometry. Sine of an angle is obtained using the SOH of SOHCAHTOA formula of trigonometry. SOH is an acronym for sine, opposite and hypotenuse. Sine of an angle is obtained from the ratio of opposite and hypotenuse of a given triangle. Thus, SOH

is denoted as stated below.

sine \:  \beta  =  \frac{opposite}{hypotenuse}

Hence, given the above triangle,

sine \: of \: angle \: u =  \frac{ |bc| }{ |ab| }  =  \frac{55}{73}

Therefore, the sine of angle u = 55/73 (D)

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