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GeometryQuestion and Answers: Page 116

Question Number 5068 Answers: 0 Comments: 0

Question Number 5051 Answers: 1 Comments: 1

7x−4=0

$7 x - 4 = 0$

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

$\frac{n^{1.2} - 3 \sqrt{3}}{n^{0.8} - \sqrt{3 n^{0.8}} + 3} - \sqrt{3} (\sqrt{3 n^{0.8}} - 1)$ $S i m p l i f y$

Question Number 5017 Answers: 0 Comments: 0

test

$t e s t$

Question Number 5030 Answers: 1 Comments: 0

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

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

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)))

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

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?

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