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GeometryQuestion and Answers: Page 111

Question Number 5068 Answers: 0 Comments: 0

Question Number 5051 Answers: 1 Comments: 1

7x−4=0

$$\mathrm{7}{x}−\mathrm{4}=\mathrm{0} \\ $$

Question Number 5022 Answers: 0 Comments: 1

((n^(1.2) −3(√3))/(n^(0.8) −(√(3n^(0.8) ))+3))−(√3)((√(3n^(0.8) ))−1) Simplify

$$\frac{{n}^{\mathrm{1}.\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{3}}}{{n}^{\mathrm{0}.\mathrm{8}} −\sqrt{\mathrm{3}{n}^{\mathrm{0}.\mathrm{8}} }+\mathrm{3}}−\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}{n}^{\mathrm{0}.\mathrm{8}} }−\mathrm{1}\right) \\ $$$$ \\ $$$${Simplify} \\ $$

Question Number 5017 Answers: 0 Comments: 0

test

$${test} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 5030 Answers: 1 Comments: 0

y=((cos x)/(√(1+sin x))) y′=? y=cos^2 4x y′=?

$${y}=\frac{\mathrm{cos}\:{x}}{\sqrt{\mathrm{1}+\mathrm{sin}\:{x}}} \\ $$$${y}'=? \\ $$$${y}=\mathrm{cos}^{\mathrm{2}} \:\mathrm{4}{x} \\ $$$${y}'=? \\ $$

Question Number 4971 Answers: 2 Comments: 7

Question Number 4847 Answers: 0 Comments: 0

(√(6+(√(6+(√(6+(√(6+(√6))))))))) SOLUTION let x = (√(6+(√(6+(√(6+(√(6+(√6))))))))) therefore.. x^(2 ) = 6+(√(6+(√(6+(√(6+(√(6 )))))))) the equation is a continuos funtion Thus x^2 = 6+(√(6+(√(6+(√(6+(√(6+(√6) ))))))))...... since x = (√(6+(√(6+(√(6+(√(6+(√6))))))))) Therdfore x^2 = 6 + x x^2 − x − 6 = 0 x^2 − 3x + 2x − 6 = 0 (x^2 − 3x) + (2x − 6) = 0 x(x − 3) + 2(x − 3) = 0 (x − 3)(x + 2) = 0 x − 3 = 0 or x − 2 = 0 x = 3 or x = −2 since negative is not allowed Thus x = 6 Meaning that (√(6+(√(6+(√(6+(√(6+(√(6 )))))))))) = 3 DONE THANK YOU SO MUCH. I UNDERSTAND THE SOLUTION.

Question Number 4783 Answers: 0 Comments: 1

cos α+β ≈(((cos α+cos β)/(cos^(−1) α+cos^(−1) β)))^(α+β) sin a+b≈(((sin a+sin b)/(sin^(−1) a+sin^(−1) b)))^(a+b) tan (a+(a/b))^k ≈(((tan (a+b)×k)/(tan^(−1) (a+b)×k)))

$$\mathrm{cos}\:\alpha+\beta\:\approx\left(\frac{\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta}{\mathrm{cos}^{−\mathrm{1}} \alpha+\mathrm{cos}^{−\mathrm{1}} \beta}\right)^{\alpha+\beta} \\ $$$$\mathrm{sin}\:{a}+{b}\approx\left(\frac{\mathrm{sin}\:{a}+\mathrm{sin}\:{b}}{\mathrm{sin}^{−\mathrm{1}} {a}+\mathrm{sin}^{−\mathrm{1}} {b}}\right)^{{a}+{b}} \\ $$$$\mathrm{tan}\:\left({a}+\frac{{a}}{{b}}\right)^{{k}} \approx\left(\frac{\mathrm{tan}\:\left({a}+{b}\right)×{k}}{\mathrm{tan}^{−\mathrm{1}} \left({a}+{b}\right)×{k}}\right) \\ $$

Question Number 4730 Answers: 0 Comments: 1

∫_0 ^(π/2) (dx/(sinx^(cosx ) + cosx^(sinx ) ))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{{sinx}^{{cosx}\:} \:\:+\:\:\:\:{cosx}^{{sinx}\:} } \\ $$

Question Number 4655 Answers: 0 Comments: 1

Question Number 4612 Answers: 0 Comments: 3

Question Number 4594 Answers: 1 Comments: 0

∫4x^2 − (5/(2x^(−2) )) + 4 dx, x≠0

$$\int\mathrm{4}{x}^{\mathrm{2}} \:−\:\frac{\mathrm{5}}{\mathrm{2}{x}^{−\mathrm{2}} }\:+\:\mathrm{4}\:\:{dx},\:{x}\neq\mathrm{0} \\ $$$$ \\ $$

Question Number 4579 Answers: 1 Comments: 2

In a right triangle, the mid-point of the hypotenuse is equidistant from all the three vertices of the triangle.

$${In}\:{a}\:{right}\:{triangle},\:{the}\:{mid}-{point}\:{of}\:{the} \\ $$$${hypotenuse}\:{is}\:{equidistant}\:{from}\:{all}\:{the} \\ $$$${three}\:{vertices}\:{of}\:{the}\:{triangle}. \\ $$

Question Number 4578 Answers: 1 Comments: 2

The segment between the mid-points of two sides of a triangle is parallel to the third side and half as long.

$${The}\:{segment}\:{between}\:{the}\:{mid}-{points} \\ $$$${of}\:{two}\:{sides}\:{of}\:{a}\:{triangle}\:{is}\:{parallel} \\ $$$${to}\:{the}\:{third}\:{side}\:{and}\:{half}\:{as}\:{long}.\: \\ $$

Question Number 4543 Answers: 1 Comments: 1

Triangle ABC has midpoints D, E and F. By connecting each verticie with the opposite midpoint, we create a cress−section called G. Prove that all three lines cross at point G regardless of the type of triangle

$$\mathrm{Triangle}\:{ABC}\:\mathrm{has}\:\mathrm{midpoints} \\ $$$${D},\:{E}\:\mathrm{and}\:{F}. \\ $$$$ \\ $$$$\mathrm{By}\:\mathrm{connecting}\:\mathrm{each}\:\mathrm{verticie}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{opposite}\:\mathrm{midpoint},\:\mathrm{we}\:\mathrm{create}\:\mathrm{a}\:\mathrm{cress}−\mathrm{section} \\ $$$$\mathrm{called}\:{G}. \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{all}\:\mathrm{three}\:\mathrm{lines}\:\mathrm{cross}\:\mathrm{at}\: \\ $$$$\mathrm{point}\:{G}\:\mathrm{regardless}\:\mathrm{of}\:\mathrm{the}\:\mathrm{type}\:\mathrm{of} \\ $$$$\mathrm{triangle} \\ $$

Question Number 4496 Answers: 0 Comments: 1

This is a simple question but for some silly reason I can′t figure it out... 1. If I have a circle with radius r and area A, and I wish to make a new circle with n times the area, for what new value of r should be used? 2. What if I increase the area from A to A+n? What value of r would be used here?

$$\mathrm{This}\:\mathrm{is}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{question}\:\mathrm{but}\:\mathrm{for}\:\mathrm{some} \\ $$$$\mathrm{silly}\:\mathrm{reason}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{figure}\:\mathrm{it}\:\mathrm{out}... \\ $$$$ \\ $$$$\mathrm{1}. \\ $$$$\mathrm{If}\:\mathrm{I}\:\mathrm{have}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:{r}\:\mathrm{and} \\ $$$$\mathrm{area}\:{A},\:\mathrm{and}\:\mathrm{I}\:\mathrm{wish}\:\mathrm{to}\:\mathrm{make}\:\mathrm{a}\:\mathrm{new}\:\mathrm{circle}\:\mathrm{with} \\ $$$${n}\:\mathrm{times}\:\mathrm{the}\:\mathrm{area},\:\mathrm{for}\:\mathrm{what}\:\mathrm{new}\:\mathrm{value} \\ $$$$\mathrm{of}\:{r}\:\mathrm{should}\:\mathrm{be}\:\mathrm{used}? \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\mathrm{What}\:\mathrm{if}\:\mathrm{I}\:\mathrm{increase}\:\mathrm{the}\:\mathrm{area}\:\mathrm{from}\:{A}\:\mathrm{to} \\ $$$${A}+{n}?\:\mathrm{What}\:\mathrm{value}\:\mathrm{of}\:{r}\:\mathrm{would}\:\mathrm{be}\:\mathrm{used}\:\mathrm{here}? \\ $$

Question Number 4505 Answers: 1 Comments: 1