Παραμύθι της Debi Gliori για την γιορτή της μητέρας.

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Η γιορτή της μητέρας πλησιάζει και αύριο στην τάξη θα μιλήσουμε για αυτή τη γιορτή. Ετοίμασα λοιπόν σε ένα χαρτί μέτρου μια αφίσα με μια γιγάντια καρδιά πάνω του. Η καρδούλα είναι άδεια, αλλά αύριο τα παιδάκια θα ρωτηθούν για ποιους λόγους αγαπούν τις μανούλες τους. Οι απαντήσεις τους θα καταγραφούν μέσα σε αυτή την καρδούλα. Όταν ένα - ένα παιδί μιλήσει, θα σηκώνεται με τη σειρά και γύρω από την καρδιά θα γράφει τη λέξη "ΜΑΜΑ" όπως εκείνο μπορεί. Αύριο που η καρδούλα δεν θα είναι πια η ίδια,.. Θα γεμίσει λόγια αγάπης! Αύριο λοιπόν, που θα συμπληρωθεί, θα την ανεβάσω! ~ ~ ~ Η καρδούλα γέμισε με υπέροχα και όμορφα λογάκια από τα παιδιά.. Οι απαντήσεις τους καταγράφηκαν πάνω στην καρδιά... δείτε τες! Και στο τέλος βρήκε τη θέση της πάνω στο ταμπλό της παρεούλας μας! Την ιδέα την άντλησα από το 3ο Νηπιαγωγείο Γέρακα (http://3nip-gerak.att.sch.gr/ekdil-11-12-giorti-miteras.htm)

Pasqual Sébah was one of the most important professional photographers of his time in the Ottoman Empire. Today, his works are highly sought after by museums and collectors. Of Syrian or Lebanese origins, Sébah (1823-1886) was a leading photographer in Constantinople, now the city of Istanbul. In 1857 he opened a studio, which he called "El Chark," next to the Russian Embassy on the Grande Rue de Pera, the main shopping street of the European part of the city. He sold photographs of the city, ancient ruins in the surrounding area, portraits, and images of local people in traditional costumes to tourists. His prints are signed P. Sébah. Sébah rose to international prominence because of his well-organized compositions, careful lighting, effective posing, attractive models, and great attention to detail. His career coincided with intense Western European interest in the "Orient," which was viewed as exotic and fascinating. In 1860, he secured the collaboration of the French photographer A. Laroche to direct his studio. As Sébah's technician, Laroche turned out photographic prints of superior quality. Sébah's career was accelerated through his collaboration with the artist, Osman Hamdi Bey (1842-1910). Osman Hamdi Bey posed models, often dressed in elaborate costumes, for Sebah to photograph. The painter then used Sébah's photographs for his celebrated Orientalist oil paintings. In 1873, Osman Hamdi Bey was appointed by the Ottoman court to direct the Ottoman exhibition in Vienna and commissioned Sébah to produce large photographs of models wearing costumes for a sumptuous album, "Les Costumes Populaires de la Turquie." The album earned Sébah a gold medal awarded by the Viennese organizers, and another medal from the Ottoman Sultan Abdulaziz. In that same exceptional year of 1873, Sébah opened a branch in Cairo, Egypt, on the Esbekieh next to the French Embassy, where he installed his associate, Laroche. (The Cairo studio remained in business until 1898.) In 1883, Sébah suffered a stroke. He died on June 15, 1886, and his brother Cosmi managed the business until Johannes (1872-1947), Pascal’s son, was old enough to take over. Johannes (Jean) grew up to become a talented photographer in his own right, but to profit from his father’s fame, he signed his photographs J. Pascal Sébah. In 1888 he went into partnership with a French photographer resident in Istanbul, Polycarpe Joaillier. The firm of Sébah and Joaillier were named the official photographers of the Sultan, and at his command took photographs all over his empire. Joaillier returned to Paris in the early 1900s, but Jean Sébah continued the studio, forming a partnership in 1910 with Hagop Iskender and Leo Perpignani. The latter left the firm in 1914. Jean Sébah and Hagop Iskender retired in 1934, leaving the business to Iskender's son, Bedros Iskender and his partner, Ismail Insel. Ismail Insel eventually became sole partner and renamed the studio Foto Sabah, which remained in business until 1952. (Sabah means "morning" in Turkish.) With all the changes, the studio that Pascal Sébah began in 1857 lasted 95 years.

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Η ταινία, "Σ΄αγαπώ" προβλήθηκε τη σαιζόν 1971-1972 και έκοψε 393.137 εισιτήρια. Ήρθε στην 2η θέση σε 90 ταινίες.

Kassos / S. Aegean

1.A + B +C = 30cm, A + B = 20 Cm. Find B + c ? 2.If we add 10 to twice a number we get 50. Find the number 3.If we subtract 30 from three times a number, we get 100. Find the number. 4.Find a number which when multiplied by 8 and then reduced by 4 is equal to 64. 5. Sum of two numbers is 100. If one exceeds the other b 5, find the numbers. 6.Sum of three consecutive integers is 27. Find the integers. 7.Find x length 5x+4 and breath x-4 of the rectangle if its perimeter is 72m. 8. After 20 year. Rama’s age will become four times that of he present age. Find he present age. 9.length of a rectangle exceeds its breadth by 5m. If the perimeter of the rectangle is 100m. Find its length and breadth. 10.Rama Said “multiplying my number by 10 and adding 20 to it gives the same answer subtracting my number from 40. Find Rama’s number. 11. A number is divided into two parts such that one part is 20 more than the other. if the two parts in the ration 4: 5 Find the number and the two parts. 12.A sum of Rs. 5000 is to be given in the from of 75 prizes. If the prize money is either Rs 100 or Rs. 25. Find the number of prizes of each type. 13. x0 + 2x0 + 3x0 = 180 Find three angles 14. I am a number Tell my identity Take me two times over and add a forty six To reach a century You still need five. 15.The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus10. The highest mark is 90. What is the lowest mark? Om Sri Ram Om Sri Ram Om Sri Ram 1.A + B +C = 30cm, A + B = 20 Cm. Find B + c ? Ans. 20 + c = 30, c= 30-20 = 10 2.If we add 10 to twice a number we get 50. Find the number Ans. Let the number = x, The twice number = 2x An adding 1= = 2x + 10 By problem: 2x + 10 = 50, 2x = 40 There fore x = 20 3.If we subtract 30 from three times a number, we get 90. Find the number. Ans. Let the number = x, Then The three times Number = 3x On subtracting 30 = 3x – 30 By problem: 3x – 30 =100, 3x = 90 – 30 = 60 Ther fore x = 60/3 = 20 4.Find a number which when multiplied by 8 and then reduced by 4 is equal to 68. Sol. Let the number = x, Then The multiply by 8 = 8x By problem: 8x – 4 =68, 8x = 68 +4 = 72, x = 72/8 =9 5. Sum of two numbers is 100. If one exceeds the other by 6, find the numbers. Sol. Lat the one number be=x, Then the other number =x-6 Sum of the numbers: x + x-6 = 2x - 6 By problem: 2x – 6 = 100, 2x = 100 + 6=106, x= 53 6.Sum of three consecutive integers is 27. Find the integers. Let the three integers be x, x+1, x+2 By problem+ 3x + 3=27, 3x = 27 -3= 24 x= 8 Therefore consecutiveintegers= 8,9,10 7.Find x length 5x+4 and breath x-4 of the rectangle if its perimeter is 72m. Sol: P=2(L+B)= 2(5x + 4 + x – 4)= 12x By problem: 12x = 72 , x = 72/12 =6 8. After 18 year. Rama’s age will become four times that of he present age. Find he present age. Let the present age of Rama be x years After 18 years Rama’s age = 4x By problem : x + 18 = 4x There fore 3x = 18, x= 6 9.langth of a rectangle exceeds its breadth by 5m. If the perimeter of the rectangle is 100m. Find its length and breadth. Sol. Breath = x, length = x + 5 P= 2 (l + b) = 2(x + 5 +X) , 4x + 10 =100 There fore 4x = 90 x=90/4 =22.5 10.Rama Said “multiplying my number by 10 and adding 20 to it gives the same answer subtracting my number from 40. Find Rama’s number. Sol. Let Rama’s number be x Multiplying by 10 and adding 20 to that number = 1ox +20 Subtracting that number 40 =40 –x By proble above two are equal i.e = 10x + 20 = 40 – x , 11x = 20, x = 20/11 11. A number is divided into two parts such that one part is 20 more than the other. if the two parts in the ration 4: 5 Find the number and the two parts. Sol. Let one part be x, Then the other part = x + 20 Ratio of these two parts= x + 20 : x = 4 : 5 X + 20 /x = 4/5, 5(x + 20)=4x, 5x + 100 = 4x, x = -100 12.A sum of Rs. 5000 is to be given in the from of 75 prizes. If the prize money is either Rs 100 or Rs. 25. Find the number of prizes of each type. Sol. Let the number of Rs. 100 prizes be x The number of Rs. 25 prizes be = 75-x , Value of the prizes = 100 x + (75-x)25 = 100x -25x+1875=0 By problem: 75x +1875 = 5000 :75x = 5000 – 1875 = 3125 X = 3125/75 = 41.6 13. x0 + 2x0 + 3x0 = 180 Find three angles 6x = 180, x = 180/6 = 30 The angles 300, 600, 900 14. I am a number Tell my identity Take me two times over and add a forty six To reach a century You still need five. Let the number be x Two times the number = 2x On adding 46 = 2x + 46 BY problem: 2x + 46 = 100-5 2x = 95-46= 49 x= 49/2 = 24.5 15.The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus10. The highest mark is 90. What is the lowest mark? Let the lowest marks of the class = x Twice least marks = 2x , on adding 2x + 10 By problem : 2x + 10 = 90, 2x = 80 , x = 40

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Παραμύθι της Debi Gliori για την γιορτή της μητέρας.