http://m2suidhat.blogspot.com/

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Mathematics Sport

𝟏𝟓𝟏

𝟐+ 𝟐

𝟐

𝟑 − 𝟗

𝟏

𝟒

𝟑𝟏

𝟐+

𝟖

𝟑 −

𝟑𝟕

𝟒

= 𝟏𝟖𝟔

𝟏𝟐+

𝟑𝟐

𝟏𝟐 −

𝟏𝟏𝟏

𝟏𝟐

= 𝟐𝟏𝟖

𝟏𝟐 −

𝟏𝟏𝟏

𝟏𝟐

= 𝟏𝟎𝟕

𝟏𝟐

= 8𝟏𝟏

𝟏𝟐

36 8

x 24

𝒙

𝟑𝟔=

𝟖

𝟐𝟒 x = 𝟑𝟔 ×

𝟖

𝟐𝟒 x =

𝟑𝟔

𝟑

x = 12 buah

𝟏𝟐𝟓𝟐𝟑 = 𝟓𝟑

𝟐𝟑 = 𝟓𝟐 = 𝟐𝟓

𝟔𝟎 : 𝟓 𝟏𝟐× 𝟓 : 𝟓

= 𝟏𝟐× 𝟓 : 𝟓

= 𝟏𝟐

= 𝟒 × 𝟑

= 2 𝟑

𝟔𝟎 : 𝟓 𝟔𝟎:𝟓

= 𝟏𝟐

= 𝟒 × 𝟑

= 2 𝟑

atau

Paket 1

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𝟐

𝟑

𝟐

𝟑 ×

𝟑

𝟑 =

𝟐 𝟑

𝟑

(i) Besar bunga = 920.000 – 800.000 = 120.000

(ii) Besar bunga = 𝒃𝒖𝒍𝒂𝒏

𝟏𝟐× 9% × 800.000

Sehingga: 120.000 = 𝒃𝒖𝒍𝒂𝒏

𝟏𝟐 ×

𝟗

𝟏𝟎𝟎 × 800.000 120 =

𝒃𝒖𝒍𝒂𝒏

𝟒× 3 × 8

bulan = 𝟏𝟐𝟎

𝟑×𝟐 = 20 bulan

b = 𝑼𝟕−𝑼𝟐

𝟕−𝟐 =

𝟑𝟏−𝟔

𝟓 =

𝟐𝟓

𝟓 = 5

U2 = a + (2-1)5

6 = a + 5

a = 1

U40 = a + (40 – 1)5

U40 = 1 + (39)5

U40 = 1 + 195

U40 =196

b = 𝑼𝟕−𝑼𝟑

𝟕−𝟑 =

𝟑𝟖−𝟏𝟖

𝟒 =

𝟐𝟎

𝟒 = 5

U3 = a + (3-1)5

18 = a + 10

a = 8

S24 = 𝟐𝟒

𝟐(2×8 + (24 – 1)5)

S24 = 12(16 + (23)5)

S24 = 12(16 + 115)

S24 = 12(131)

S24 =1572

Barisan gaji: 3.000.000, 3.500.000, 4.000.000, 4.500.000,....

S10 = 𝟏𝟎

𝟐(2×3.000.000 + (9)500.000)

S10 = 5(6.000.000 + 4.500.000)

S10 = 5(10.500.000)

S10 = 52.500.000

(i) 15x2y – 20xy

2 = 5xy(3x – 4y)

(ii) p2 – 16 = (p + 4)(p – 4)

(iii) 3a2 + 8a – 3 = (3a – 1)(a + 3)

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3x + 6 = 5x + 20

6 – 20 = 5x – 3x

– 14 = 2x

– 7 = x

x = – 7

K = 2(p + l)

44 = 2(p + l)

22 = (p + l)

22 = (3x + 4) + (2x + 3)

22 = 5x + 7

22 – 7 = 5x

15 = 5x

3 = x

x = 3

p = 3x + 4 = 3(3) + 4 = 9 + 4 = 13

l = 2x + 3 = 2(3) + 3 = 6 + 3 = 9

Banyak anggota D = 6

Banyak himpunan bagian dari D adalah 26 = 64

P C 40

12

23

-12

= 1

1

40 – (11 + 12) = 40 – 23 = 17

Menulis cerpen = 17 + 12 = 29

f(x) = 8 – 2x f(k) = 8 – 2k

–10 = 8 – 2k

–18 = –2k

9 = k

k = 9

Jadi, x + 12 = – 7 + 12 = 5

Paket 1

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m1 = 𝟐−𝟕

𝟎+𝟏 = – 5

m1×m2 = –1

–5 × m2 = –1

m2 = 𝟏

𝟓

m = 2

Memotong sumbu-y di titik (0, 2)

m = −𝟑

𝟐

Memotong sumbu-y bukan di titik (0, 2) m =

𝟐

𝟏 = 2

Memotong sumbu-y bukan di titik (0, 2)

m = 𝟏

𝟏 = 1

Memotong sumbu-y bukan di titik (0, 2)

m = 𝟐

𝟏 = 2

Memotong sumbu-y di titik (0, 2)

A. m2 = 5

B. m2 = 𝟏

𝟓

C. m2 = – 5

D. m2 = – 5

m1 = −𝟏𝟑−𝟏

−𝟒−𝟑 =

−𝟏𝟒

−𝟕 = 2

y – y1 = m(x – x1), (ambil x1 = 3 dan y1 = 1)

y – 1 = 2(x – 3)

y – 1 = 2x – 6

y = 2x – 5

y = 2x – 5 , titik A (10, p)

p = 2(10) – 5

p = 20 – 5

p = 15

x – 3y – 5 = 0 ×2 2x – 6y – 10 = 0

2x – 5y – 9 = 0 ×1 2x – 5y – 9 = 0

–y – 1 = 0

y = –1 sehingga x – 3y – 5 = 0

x – 3(–1) – 5 = 0

x + 3 – 5 = 0

x – 2 = 0

x = 2

Nilai 3x + 2y = 3(2) + 2(–1)

= 6 – 2

= 4

Paket 1

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5a + 3j = 79.000 ×2 10a + 6j = 158.000

3a + 2j = 49.000 ×3 9a + 6j = 147.000

a = 11.000

Panjang tali = 𝟏𝟓𝟎𝟐 + 𝟏𝟓𝟎𝟐 = 𝟐 × 𝟏𝟓𝟎𝟐 =𝟏𝟓𝟎 𝟐

Panjang perkiraan dengan 𝟐 𝟏,𝟒𝟏 adalah 150 × 1,41 = 211,5 212

K L

N M

B

C

D

A

K L

N M

B

CD

A cm8

cm8

cm10

cm10

cm10

Luas daerah yang diarsir = 𝟏

𝟒 × Luas persegi KLMN

= 𝟏

𝟒 × (8)

2

= 16 cm2

Paket 1

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(4 + 4 + 6 + 6) × 2 = (20) × 2 = 40 cm

4 cm

4 cm

6 cm

6 cm

atau

1

2

4

3

(i) 1 dengan 3

(ii) 2 dengan 4

(iii) 12 dengan 34

(iv) 14 dengan 23

ada 4 pasang

x

Q

R

P

S

T

R

10 cm

15 cm

6 cm

𝑻𝑹

𝑸𝑹=

𝑹𝑺

𝑷𝑹

𝑻𝑹

𝟏𝟎=

𝟔

𝟏𝟓

TR = 𝟏𝟎×𝟔

𝟏𝟓

TR = 4 cm

Jadi, PT = PR – TR

= 15 – 4

PT = 11 cm

Paket 1

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D

A

E

B

C

E

x

x

𝑨𝑫

𝑩𝑪=𝑨𝑬

𝑬𝑪=𝑫𝑬

𝑩𝑬

A + B = 1800

(2x + 30) + (5x + 10) = 180

7x + 40 = 180

7x = 140

x = 20

K

L

M

4

1

2

3

Jadi, B = (5x + 10) = 5(20) + 10

= 100 + 10

B = (5x + 10) = 1100

K

L

M

3

(i)

K

L

M

2

(ii)

K

L

M

1

(iii)

K

L

M

4

(iv)

Panjang busur AB = 𝟕𝟐𝐨

𝟑𝟔𝟎𝐨 × 2 r =

𝟏

𝟓 × 2 ×

𝟐𝟐

𝟕 × 14

= 𝟏

𝟓 × 44 × 2

= 𝟖𝟖

𝟓

= 17, 6 cm

Paket 1

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d2 = p

2 – (r1 + r2)

2 p

2 = d

2 + (r1 + r2)

2

= 162 + (8 + 4)

2

= 256 + (12)2

= 256 + 144

= 400

p = 20

Ada 18 (6×3) rusuk dan 8 (6 + 2) sisi

V = La × t

= 𝟏𝟖+𝟏𝟐

𝟐× 𝟏𝟎 × 20

= (30 × 5) × 20

= 150 × 20

V = 3000

Dengan menggunakan teorema pythagoras

didapat panajang TE = 10 cm

L = luas alas + 4 × luas sisi tegak

= 122 + 4 ×

𝟏

𝟐 × 12 × 10

= 144 + 240

L = 384 cm2

12 cm

8 cm

E

O

Paket 1

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LP = Luas tabung tanpa tutup + luas setengah bola

= (πr2 + 2πrt) + 2πr

2

= (𝟐𝟐

𝟕×7

2 + 2×

𝟐𝟐

𝟕×7×20) + 2×

𝟐𝟐

𝟕×7

2

= (22×7 + 2×22×20) + 2×22×7

LP = 1.342

4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10

Median = 𝟔+𝟕

𝟐 = 6,5

Bisa jadi, rata-ratanya = 130

Bisa jadi, rata-ratanya = 130

Bisa jadi, rata-ratanya = 130

Bisa dipastikan, rata-ratanya = 130

Rata-rata = 𝟐𝟑×𝟏𝟑𝟎+𝟏𝟑𝟑+𝟏𝟐𝟕

𝟐𝟑+𝟏+𝟏 =

𝟑𝟐𝟓𝟎

𝟐𝟓 = 130

180 130 525 640 360 200 2035

30 Rata-rata =

𝟐𝟎𝟑𝟓

𝟑𝟎 = 67, 83

Banyak siswa yang nilainya lebih dari rata-rata = 30 – 9 = 21

Paket 1

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Senin – Selasa dan Kamis – Jum’at penurunan 1000

Selasa – Rabu kenaikan 2000

Rabu – Kamis dan Jum’at – Sabtu uang saku tetap

Senin – Selasa penurunan 1000

Nomor bola kurang dari 5 1, 2, 3, 4 ada 4

P = 𝒏𝒐𝒎𝒐𝒓 𝒃𝒐𝒍𝒂 𝒌𝒖𝒓𝒂𝒏𝒈 𝒅𝒂𝒓𝒊 𝟓

𝒃𝒂𝒏𝒚𝒂𝒌 𝒏𝒐𝒎𝒐𝒓 𝒃𝒐𝒍𝒂 =

𝟒

𝟗

Disusun oleh : Mohammad Tohir

Jika ada saran, kritik maupun masukan

silahkan kirim ke- My email: mohammadtohir@yahoo.com

Terima kasih.

My blog : http://m2suidhat.blogspot.com/

http://matematohir.wordpress.com/

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