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Q 50+1. 1 + 2 + 3 + 4 +

Berikut ini adalah pertanyaan dari anitasamosir17 pada mata pelajaran Matematika untuk jenjang Sekolah Dasar

Q 50+1. 1 + 2 + 3 + 4 + 5 + ... + 21 =

2. Sebuah balok mempunyai panjang 15 cm Lebar 12 cm Tinggi 12 cm.
Tentukan volume dan Luas permukaannya

3. 7! × 10 × 3! = ​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

#1

Sn = n/2 × (2a + (n - 1)b)

a → 1

b → 1

n → 21

S21 = 21/2 × (2(1) + (21 - 1)1)

S21 = 10,5 × (2 + (20)1)

S21 = 10,5 × (2 + 20)

S21 = 10,5 × 22

S21 = 231

#2

V = P × L × T

V = 15 × 12 × 12

V = 15 × 144

V = 2.160 cm³

Lp = 2 × (PL + PT + LT

Lp = 2 × (15.12 + 15.12 + 12.12)

Lp = 2 × (180 + 180 + 144)

Lp = 2 × (180 + 324)

Lp = 2 × 504

Lp = 1.008 cm²

#3

7! × 10 × 3!

= (7.6.5.4.3.2.1) × 10 × (3.2.1)

= (42.5.4.3.2.1) × 10 × (6.1)

= (210.4.3.2.1) × 10 × 6

= (840.3.2.1) × 10 × 6

= (2.520.2.1) × 10 × 6

= (5.040.1) × 10 × 6

= 5.040 × 10 × 6

= 5.040 × 60

= 302.400

:)

#1Sn = n/2 × (2a + (n - 1)b)a → 1b → 1n → 21S21 = 21/2 × (2(1) + (21 - 1)1)S21 = 10,5 × (2 + (20)1)S21 = 10,5 × (2 + 20)S21 = 10,5 × 22S21 = 231 #2V = P × L × TV = 15 × 12 × 12V = 15 × 144V = 2.160 cm³ Lp = 2 × (PL + PT + LTLp = 2 × (15.12 + 15.12 + 12.12)Lp = 2 × (180 + 180 + 144)Lp = 2 × (180 + 324)Lp = 2 × 504Lp = 1.008 cm² #37! × 10 × 3!= (7.6.5.4.3.2.1) × 10 × (3.2.1)= (42.5.4.3.2.1) × 10 × (6.1)= (210.4.3.2.1) × 10 × 6= (840.3.2.1) × 10 × 6= (2.520.2.1) × 10 × 6= (5.040.1) × 10 × 6= 5.040 × 10 × 6= 5.040 × 60= 302.400 :)#1Sn = n/2 × (2a + (n - 1)b)a → 1b → 1n → 21S21 = 21/2 × (2(1) + (21 - 1)1)S21 = 10,5 × (2 + (20)1)S21 = 10,5 × (2 + 20)S21 = 10,5 × 22S21 = 231 #2V = P × L × TV = 15 × 12 × 12V = 15 × 144V = 2.160 cm³ Lp = 2 × (PL + PT + LTLp = 2 × (15.12 + 15.12 + 12.12)Lp = 2 × (180 + 180 + 144)Lp = 2 × (180 + 324)Lp = 2 × 504Lp = 1.008 cm² #37! × 10 × 3!= (7.6.5.4.3.2.1) × 10 × (3.2.1)= (42.5.4.3.2.1) × 10 × (6.1)= (210.4.3.2.1) × 10 × 6= (840.3.2.1) × 10 × 6= (2.520.2.1) × 10 × 6= (5.040.1) × 10 × 6= 5.040 × 10 × 6= 5.040 × 60= 302.400 :)#1Sn = n/2 × (2a + (n - 1)b)a → 1b → 1n → 21S21 = 21/2 × (2(1) + (21 - 1)1)S21 = 10,5 × (2 + (20)1)S21 = 10,5 × (2 + 20)S21 = 10,5 × 22S21 = 231 #2V = P × L × TV = 15 × 12 × 12V = 15 × 144V = 2.160 cm³ Lp = 2 × (PL + PT + LTLp = 2 × (15.12 + 15.12 + 12.12)Lp = 2 × (180 + 180 + 144)Lp = 2 × (180 + 324)Lp = 2 × 504Lp = 1.008 cm² #37! × 10 × 3!= (7.6.5.4.3.2.1) × 10 × (3.2.1)= (42.5.4.3.2.1) × 10 × (6.1)= (210.4.3.2.1) × 10 × 6= (840.3.2.1) × 10 × 6= (2.520.2.1) × 10 × 6= (5.040.1) × 10 × 6= 5.040 × 10 × 6= 5.040 × 60= 302.400 :)#1Sn = n/2 × (2a + (n - 1)b)a → 1b → 1n → 21S21 = 21/2 × (2(1) + (21 - 1)1)S21 = 10,5 × (2 + (20)1)S21 = 10,5 × (2 + 20)S21 = 10,5 × 22S21 = 231 #2V = P × L × TV = 15 × 12 × 12V = 15 × 144V = 2.160 cm³ Lp = 2 × (PL + PT + LTLp = 2 × (15.12 + 15.12 + 12.12)Lp = 2 × (180 + 180 + 144)Lp = 2 × (180 + 324)Lp = 2 × 504Lp = 1.008 cm² #37! × 10 × 3!= (7.6.5.4.3.2.1) × 10 × (3.2.1)= (42.5.4.3.2.1) × 10 × (6.1)= (210.4.3.2.1) × 10 × 6= (840.3.2.1) × 10 × 6= (2.520.2.1) × 10 × 6= (5.040.1) × 10 × 6= 5.040 × 10 × 6= 5.040 × 60= 302.400 :)

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Last Update: Fri, 30 Sep 22
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