Because two segments AB and CD are symmetric to each other with respect to axis d

=> AB = CD = 3cm

So CD = 3cm

5/89 + 8/89 = 13/89 

\(A=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{123}-\dfrac{1}{126}\right)\)

\(A=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{126}\right)=\dfrac{1}{3}\cdot\dfrac{41}{126}=\dfrac{41}{378}\)

So \(A=\dfrac{41}{378}\)

Because ABC is the isosceles triangles

We have: \(\widehat{A}\) = 90^o

 * If \(\widehat{A}\) is a corner side => \(\widehat{A}=\widehat{B}or\widehat{C}=90^o\)

=> Last corner = 0^o [nosatisfied]

 * If \(\Delta ABC\) isosceles in A

=> \(\widehat{B}=\widehat{C}=\dfrac{180^0-90^0}{2}=45^0\) [satisfied]

So \(\widehat{B}=\widehat{C}=45^0\)

We have:

\(\widehat{A_1}+\widehat{D_2}+\widehat{C}=180^o\)

\(\widehat{A_2}+\widehat{D_1}+\widehat{B}=180^o\)

\(\Rightarrow\widehat{A_1}+\widehat{D_2}+\widehat{C}+\widehat{A_2}+\widehat{D_1}+\widehat{B}=180^0+180^0\)

=> \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

So the sum of four angles in a quadrilateral is 360^o

1 + 2 + 2² + ... + 2¹⁴

= [1 + 2 + 2² + 2³ + 2⁴] + 2⁶[1 + 2 + 2² + 2³ + 2⁴] + 2¹¹[1 + 2 + 2² + 2³]

= 31 + 2⁶.31 + 2¹¹.15 

But 2¹¹.15 \(⋮̸31\)

=> A \(⋮̸31\)

This question was wrong

a,

x² + x + 1

= x² + 2.x.1/2 + [1/2]² + 3/4

= [x + 1/2]² + 3/4 \(\ge\) 3/4 

=> x² + x + 1 > 0

b,

2x² + 2x + 1

= x² + x² + 2.x.1 + 1²

= x² + [x+1]² \(\ge\)0 + 0 = 0

=> 2x² + 2x + 1 \(\ge0\)

A= 1/3[1/3 - 1/6 + 1/6 - 1/9 +... + 1/132 - 1/135]

= 1/3[1/3 - 1/135]

= 1/3*44/135

= 44/405

= 862.38 + 862.62

= 862.\(\left(38+62\right)\)

= 862.100

= 86200

Help you solve math was wrong

\(A=\dfrac{1}{6\cdot10}+\dfrac{1}{10\cdot14}+...+\dfrac{1}{202\cdot206}\)

\(A=\dfrac{1}{4}\left(\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{202}-\dfrac{1}{206}\right)\)

\(A=\dfrac{1}{4}\left(\dfrac{1}{6}-\dfrac{1}{206}\right)\)

\(A=\dfrac{1}{4}\cdot\dfrac{50}{309}\)

\(A=\dfrac{25}{618}\)

So \(A=\dfrac{25}{618}\)

With x = -1

=> x¹⁰⁰ + x⁹⁹ + x⁹⁸ + ... + x + 1

= \(\left(-1\right)^{100}+\left(-1\right)^{99}+\left(-1\right)^{98}+...+\left(-1\right)+1\)

= \(1+\left(-1\right)+1+...+\left(-1\right)+1\)

\(=\left[1+\left(-1\right)\right]+\left[1+\left(-1\right)\right]+...+\left[1+\left(-1\right)\right]+1\)

\(=0+0+...+0+1=1\)

Because Daniella ate more cookies thann any of other kids 

=> The cookies of Daniella > 3 

But have 20 cookies

Again: 1 + 2 + 3 + x + y + z = 20

=> x + y + z = 14

Tell the cookies of Daniella is z and z > x, z > y

=> x + y is maximun so z is minimun and z > x, z > y

=> x + y = 8 and x = y = 4 or x = 3 and y = 5 and repeat.

=> z = 6

So the smallest possible number of cookies that Daniella ate is 6 cookies

Tell the numbers of this answer is a

=> a\(⋮3,4,5\) and \(0\le a\le499\)

We have:

3 = 3

4 = 2²

5 = 5

=> \(LCM\left(3,4,5\right)=2^2.3.5=60\)

=> The multiples of all numbers 3, 4 and 5 is multiples of 60

=> a \(\in\left\{0,60,120,180,240,300,360,420,480,...\right\}\)

But \(0\le a\le499\)

=> a \(\in\left\{0,60,120,180,240,300,360,420,480\right\}\)

So there are 9 numbers

50 + 300 + 50 + 100

= [50 + 50] + 300 + 100

= 100 + 300 + 100

= 500

x : 2 + x + x:3 + x: 4 = 25

=> x/2 + x/1 + x/3 + x/4 = 25

=> 6x/12 + 12x/12 + 4x/12 + 3x/12 = 15

=> \(\dfrac{6x+12x+3x+4x}{12}=25\)

\(\Rightarrow\dfrac{25x}{12}=25\)

=> 25x = 25.12

=> x = 12

70 + 12 = 82

56 + 82 = 138

60 + 12 = 72

I will tk you

90 + 501 = 591