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GeometryQuestion and Answers: Page 114

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)))

$\cos α + β \approx {(\frac{\cos α + \cos β}{\cos^{- 1} α + \cos^{- 1} β})}^{α + β}$ $\sin a + b \approx {(\frac{\sin a + \sin b}{\sin^{- 1} a + \sin^{- 1} b})}^{a + b}$ $\tan {(a + \frac{a}{b})}^{k} \approx (\frac{\tan (a + b) \times k}{\tan^{- 1} (a + b) \times k})$

Question Number 4730 Answers: 0 Comments: 1

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

$\int_{0}^{\frac{π}{2}} \frac{d x}{{s i n x}^{c o s x} + {c o s x}^{s i n x}}$

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

$\int 4 x^{2} - \frac{5}{2 x^{- 2}} + 4 d x, x \neq 0$

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.

$I n a r i g h t t r i a n g l e, t h e m i d - p o i n t o f t h e$ $h y p o t e n u s e i s e q u i d i s t a n t f r o m a l l t h e$ $t h r e e v e r t i c e s o f t h e t r i a n g l e .$

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.

$T h e s e g m e n t b e t w e e n t h e m i d - p o i n t s$ $o f t w o s i d e s o f a t r i a n g l e i s p a r a l l e l$ $t o t h e t h i r d s i d e a n d h a l f a s l o n g .$

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

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

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?

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

Question Number 4505 Answers: 1 Comments: 1

Question Number 4437 Answers: 1 Comments: 0

An ellipse having semi-major axis length a and semi-minor axis length b and a circle having radius r have equal area. Express r in terms of a and b.

$An ellipse having semi - major axis$ $length a and semi - minor axis length b$ $and a circle having radius r have equal$ $area .$ $Express r in terms of a and b .$

Question Number 4409 Answers: 1 Comments: 1

Divide a circle into two congruent regions such that they have no straightedge.

$D i v i d e a c i r c l e i n t o t w o c o n g r u e n t$ $r e g i o n s s u c h t h a t t h e y h a v e n o$ $s t r a i g h t e d g e .$

Question Number 4390 Answers: 1 Comments: 0

Question Number 4387 Answers: 0 Comments: 3

Question Number 4384 Answers: 0 Comments: 1

A circle of radius r_1 has been divided into two parts of equal area, by an arc having center on the circle. Determine the radius(r_2 ) of the arc.

$A circle of radius r_{1} has been divided$ $into two parts of equal area,$ $by an arc having center on the circle .$ $Determine the radius (r_{2}) of the arc .$

Question Number 4374 Answers: 0 Comments: 1

Question Number 4356 Answers: 1 Comments: 1

Question Number 4329 Answers: 1 Comments: 1

Question Number 4326 Answers: 3 Comments: 0

hola

$h o l a$

Question Number 4318 Answers: 1 Comments: 2

Question Number 4268 Answers: 3 Comments: 0

1+2=

$1 + 2 =$

Question Number 4219 Answers: 1 Comments: 0

Question Number 4225 Answers: 0 Comments: 0

^• A kite is a quadrilateral having two pairs of adjacent sides equal. Draw a semi-circle inside it touching all the sides using Eucledian tools. ^• Can we show that the above semi-circle is of the laregest possible area inside the kite?

$^{∙} A kite is a quadrilateral having two$ $pairs of adjacent sides equal .$ $Draw a semi - circle inside it touching$ $all the sides using Eucledian tools .$ $^{∙} Can we show that the above semi - circle$ $is of the laregest possible area inside the$ $kite ?$

Question Number 4080 Answers: 2 Comments: 1

Prove that inside a square, a semi-circle, which touches all the four sides of the square, is possible with ruler and compass.

$Prove that inside a square, a semi - circle,$ $which touches all the four sides of the$ $square, is possible with ruler and compass .$

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