•The........ Digest part of the food •Connect the back of the thoart at the...... •produce........ To digest food •Stores undigested food until it is passed out of the body throught the...

1. •The........ Digest part of the food •Connect the back of the thoart at the...... •produce........ To digest food •Stores undigested food until it is passed out of the body throught the...

Jawaban:

kucing goyang hot

Penjelasan:

sama sama kalau benar

2. Which part of the body uses mechanical and chemical methods to digest food? A. small instine B. Colon C. stomach​

jawabannya a. small instine
maf ya klo slh

Jawaban:

A. Small instine

Penjelasan:

In the body, mechanical digestion is carried out through three processes, namely the chewing process in the mouth, stirring in the stomach, and segmentation in the small intestine.

3. Which part of the body uses mechanical and chemical methods to digest food? A. Colon B. Small instestine C. Stomach

Jawaban:

Bagian tubuh mana yang menggunakan metode mekanis dan kimiawi untuk mencerna makanan?

A. Colon B. Usus halus C. Lambung

C. Lambung / Stomach

sorry if wrong:)

4. jaundice 11th and Rainbow​

Jawaban:

kakjajsjnBjnaiAjoakahhBiajjaoaoa

5. QUIZ MATHS Terlampir

Jawab:

[tex]\displaystyle \frac{1}{2}-\frac{1}{22!}[/tex]

Penjelasan dengan langkah-langkah:

Ubah ke bentuk notasi sigma :

[tex]\displaystyle \frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+...+\frac{22}{20!+21!+22!}\\=\sum_{n=1}^{20}\frac{n+2}{n!+(n+1)!+(n+2)!}\\=\sum_{n=1}^{20}\frac{n+2}{n!+(n+1)(n)!+(n+2)(n+1)(n)!}\\=\sum_{n=1}^{20}\frac{n+2}{n!(1+(n+1)+(n+2)(n+1))}\\=\sum_{n=1}^{20}\frac{n+2}{n!((n+2)+(n+2)(n+1))}\\=\sum_{n=1}^{20}\frac{n+2}{n!(n+2)^2}\\=\sum_{n=1}^{20}\frac{1}{n!(n+2)}.....(kali\:dengan\:\frac{n+1}{n+1})[/tex]

[tex]\displaystyle =\sum_{n=1}^{20}\frac{n+1}{(n+2)!}\\=\sum_{n=1+2}^{20+2} \frac{n-2+1}{(n-2+2)!}\\=\sum_{n=3}^{22} \frac{n-1}{n!}\\=\sum_{n=3}^{22} \frac{n}{n!}-\frac{1}{n!}\\=\sum_{n=3}^{22} \frac{n}{n(n-1)!}-\frac{1}{n!}\\=\sum_{n=3}^{22} \frac{1}{(n-1)!}-\frac{1}{n!}\\=\frac{1}{(3-1)!}-\frac{1}{3!}+\frac{1}{(4-1)!}-\frac{1}{4!}+...+\frac{1}{(22-1)!}-\frac{1}{22!}\\=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{21!}-\frac{1}{22!}\\\\=\frac{1}{2}-\frac{1}{22!}[/tex]

diubah ke bentuk persamaan sigma

[tex]\displaystyle\sum_{n=3}^{22}\frac{n}{(n-2)!+(n-1)!+n!)\\ = \sum_{n=3}^{22} \frac{n}{(n-2)!(1+(n-1)+(n-1)n}[/tex]

[tex]\displaystyle = \sum_{n=3}^{22} \frac{n}{(n-2)!(n+n^2-n)}\\ = \sum_{n=3}^{22} \frac{n}{n^2(n-2)!)\\ =\sum_{n=3}^{22} \frac{1}{n(n-2)!}[/tex]

[tex]\displaystyle = \sum_{n=3}^{22}\frac{n-1}{n(n-1)(n-2)!}\\=\sum_{n=3}^{22}\frac{n-1}{n!}\\ = \sum_{n=3}^{22}\frac{n}{n!}-\frac{1}{n!}[/tex]

[tex]\displaystyle\sum_{n=3}^{22}\frac{1}{(n-1)!} - \frac{1}{n!}\\ = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} + \dots + \frac{1}{21!} - \frac{1}{22!}[/tex]

banyak yang saling menghilangkan menyisakan

[tex]\displaystyle \frac{1}{2!} + 0 + \dots + 0 - \frac{1}{22!}\\ \boxed{=\frac{1}{2}-\frac{1}{22!}}[/tex]

6. July 11th,1945 artinya​

Jawaban:juli ke 11, 1945

Penjelasan:semoga membantu

Jawaban:

11 Juli 1945

atau

Sebelas Juli Sembilan Belas Empat puluh Lima

Penjelasan:

Di tanggal ini ada Sidang BPUPKI

Semoga membantu ^ v ^

100% Kata Asli

7. QUIZ MATHS Terlampir

Jawab:

2575

Penjelasan dengan langkah-langkah:

[tex]\displaystyle {U}_{1}=1\\{U}_{2}=1+2=3\\{U}_{3}=1+2+3=6\\...\\{U}_{r}=\frac{r}{2}(r+1)[/tex]

misalkan :

[tex]\displaystyle {x}_{1}=\frac{1}{\frac{1}{{U}_{1}}}=1\\ {x}_{2}=\frac{2}{\frac{1}{{U}_{1}}+\frac{1}{{U}_{2}}}=\frac{2}{1+\frac{1}{3}}=\frac{3}{2}\\{x}_{3}=\frac{3}{\frac{1}{{U}_{1}}+\frac{1}{{U}_{2}}+\frac{1}{{U}_{3}}}=\frac{3}{\frac{4}{3}+\frac{1}{6}}=2\\...\\{x}_{n}=1+\frac{1}{2}(n-1)\\maka,\:{x}_{100}=1+\frac{99}{2}=\frac{101}{2}[/tex]

Sehingga :

[tex]\displaystyle {x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{100}=\frac{100}{2}\left({x}_{1}+{x}_{100}\right)=50\left(1+\frac{101}{2}\right)\\=50\left(\frac{103}{2}\right)=25\times103=2575[/tex]

Cara lainnya adalah menganalisa bentuk dari soalnya, apabila diubah ke notasi sigma menjadi:

[tex]\displaystyle {x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{100}=\displaystyle \sum_{N=1}^{100}\frac{N}{S}\\dengan\:\\S=\sum_{k=1}^{N} \frac{1}{\displaystyle \sum_{j=1}^{k}j}[/tex]

jumlah bilangan asli berurutan 1+2+3+...+k adalah [tex]\displaystyle \frac{k}{2}(k+1)[/tex]

Maka :

[tex]\displaystyle \sum_{k=1}^{N} \frac{1}{\displaystyle \sum_{j=1}^{k}j}\\=\sum_{k=1}^{N} \frac{1}{\displaystyle \frac{k}{2}(k+1)}=\sum_{k=1}^{N} \frac{2}{\displaystyle k(k+1)}\\=2\sum_{k=1}^{N} \frac{1}{\displaystyle k(k+1)}\\=2\sum_{k=1}^{N} \left(\frac{1}{k}-\frac{1}{k+1}\right)........\:\:Deret\:Teleskopis\\=2\left(1-\frac{1}{N+1}\right)\\S=2\left(1-\frac{1}{N+1}\right)\\\\\\Jadi, Jumlah\:deret\:pada\:soal\:adalah\:=\displaystyle \sum_{N=1}^{100}\frac{N}{\displaystyle 2\left(1-\frac{1}{N+1}\right)}\\[/tex]

Sehingga :

[tex]\displaystyle \sum_{N=1}^{100}\frac{N}{\displaystyle 2\left(1-\frac{1}{N+1}\right)}\\= \frac{1}{2}\sum_{N=1}^{100}\frac{N}{\displaystyle\left(\frac{N}{N+1}\right)}\\=\frac{1}{2}\sum_{N=1}^{100} (N+1)\\=\frac{1}{2}\times\frac{100}{2}\times(2\times2+(100-1)\times1)\\=25(4+(100-1))\\=25(103)\\=2575[/tex]

8. Tuliskan hal yang kamu ketahui tentang Message-Digest Algortihm 4?​

Jawaban:

Message-Digest algortihm 4(seri ke-4) yang dirancang oleh Profesor Ronald Rivest dari MIT pada tahun 1990. Panjangnya adalah 128 bit. MD4 juga digunakan untuk menghitung NT-hash ringkasan password pada Microsoft Windows NT, XP dan Vista. ... SHA adalah Secure Hash Algoritma.

Penjelasan:

maaf ya kalau salah

9. 1.dari kata kunci berikut yg menghasilkan pencarian file dgn jenis PDF adalah... a. filetype:PDF b. filetype=PDF c. filetype PDF d. foletype-PDF

c, file type pdf itu yang benar menurut sayasemua jawaban salah jawaban adalah filetype.PDF
jawaban c adalah yang paling benar sepertinya

10. Part 1Tolong jawab YGY masih part 1 ada lagi part 2 nyaa​

Penjelasan dengan langkah-langkah:

(6x + 2) + (2x - 9)

8x - 7 (A)

12 × 4/3a

16a (A)

-5x + 10 + 3x - 9

-2x + 1 (A)

1). (6x + 2) + (2x - 9)

= (6x + 2x) + (-9 + 2)

= 8x + (-7)

= 8x-7

================

2). 12× 4/3a

= 48/3a

= 16a

================

3). -5x + 10 + 3x - 9

= (-5x + 3x) + (10 - 9)

= -2x+1

[tex]\bold{\mathbb{\color{ff0000}{♡} \color{ff4000}{♡}\color{ff8000}{♡}\color {ffc000}{♡}\color{ffff00}{♡}\color{c0ff00}{♡}\color {80ff00}{♡}\color{40ff00}{♡}\color {00ff00}{♡}\color{00ff40}{M}\color{00ff80}{a}\color {00ffc0}{s} \: \color{00ffff}{R}\color {00c0ff}{y}\color{0080ff}{a}\color{0040ff}{n}\color {0000ff}{♡}\color{4000ff}{♡}\color{8000ff}{♡}\color{c000ff}{♡}\color{ff00ff}{♡}\color {ff00c0}{♡}\color{ff00a0}{♡}\color{ff0080}{♡}\color{ff0040}{♡}}}[/tex]

11. MATHS PROBLEM Terlampir

Misal: [tex] \displaystyle f(x) = \sqrt{x^2+4}+\sqrt{x^2-24x+153}[/tex]

[tex]\displaystyle \min\{f(x)\} = \dots?[/tex]

Penyelesaian:

Mencari turunan [tex] f(x) [/tex]

Turunan [tex]f(x)[/tex] bentuk [tex]f(x) = \sqrt{u}[/tex] adalah

[tex] \displaystyle \boxed{f'(x) = \frac{u'}{2\sqrt{u}}}[/tex]

sehingga

[tex] \displaystyle f(x) = \sqrt{x^2+4}+\sqrt{x^2-24x+153} \\ f'(x) = \frac{2x}{2\sqrt{x^2+4}}+\frac{2x-24}{2\sqrt{x^2-24x+153}} \\ f'(x) = \frac{x}{\sqrt{x^2+4}}+\frac{x-12}{2\sqrt{x^2-24x+153}} \\ f'(x) = \frac{x\sqrt{x^2-24x+153}+(x-12)\sqrt{x^2+4}}{\sqrt{\left(x^2+4\right)\left(x^2-24x+153\right)}} [/tex]

Cari titik stasioner [tex]f(x) \to f'(x) = 0 [/tex]

[tex] \displaystyle f'(x) = 0 \\ \frac{x\sqrt{x^2-24x+153}+(x-12)\sqrt{x^2+4}}{\sqrt{\left(x^2+4\right)\left(x^2-24x+153\right)}} = 0 [/tex]

Abaikan pembilang karena pembilang ≠ 0

[tex] \displaystyle x\sqrt{x^2-24x+153}+(x-12)\sqrt{x^2+4} = 0[/tex]

untuk mempersingkat, dimisalkan

[tex] \displaystyle a=x^2-24x+153 \\ b=x^2+4[/tex]

sehingga

[tex] \displaystyle \left(x\sqrt{a}+(x-12)\sqrt{b}\right)^2 = 0 \\ ax^2+2x(x-12)\sqrt{ab}+b(x-12)^2 = 0 \\ \left(ax^2+b(x-12)^2\right)^2 = \left(-2x(x-12)\sqrt{ab}\right)^2 \\ a^2x^4+2abx^2(x-12)^2+b^2(x-12)^4 = 4abx^2(x-12)^2 \\ a^2x^4-2abx^2(x-12)^2+b^2(x-12)^4 = 0 \\ \Big(ax^2-b(x-12)^2\Big)^2 = 0 \\ ax^2-b(x-12)^2 = 0 \\ ax^2-b(x^2-24x+144) = 0 \\ (a-b)x^2+24bx-144b = 0 \\ \Big(x^2-24x+153-x^2-4\Big)x^2+24x(x^2+4)-144(x^2+4) = 0 \\ (-24x+149)x^2+24x^3+96x-144x^2-576 = 0 \\ -24x^3+24x^3+149x^2-144x^2+96x-576 = 0 \\ 5x^2+96x-576 = 0 \\ (x+24)(5x-24) = 0 \\ \begin{array}{lcl}x+24=0&\text{atau}&5x-24=0 \\ x=-24&\text{atau}&x=\frac{24}{5} \\ \bold{(TM)}&{}&{} \end{array}[/tex]

Uji [tex]f'(x)[/tex] dan abaikan pembilang karena pembilang pasti selalu positif (syarat fungsi bentuk akar)

[tex] \displaystyle \begin{aligned} \{x<\frac{24}{5}\}&: x=0 \to (0)\sqrt{(0)^2-24(0)+153}+((0)-12)\sqrt{(0)^2+4} &= 0+(-) < 0 \\ \{x>\frac{24}{5}\}&: x=12 \to (12)\sqrt{(12)^2-24(12)+153}+((12)-12)\sqrt{(12)^2+4} &= (+)+0 > 0 \end{aligned}[/tex]

Dari uji titik [tex]f'(x)[/tex], ketika diilustrasikan akan seperti ini dalam bentuk garis bilangan:

[tex] \displaystyle \boxed{\:\:\:\text{turun (-)}\:\:\:}\frac{24}{5}\boxed{\:\:\:\text{naik (+)}\:\:\:}[/tex]

Dilihat dari garis bilangan [tex]f'(x)[/tex], nilai [tex]\min\{f(x)\}[/tex] didapat ketika [tex]x=\frac{24}{5}[/tex] sehingga nilai [tex]\min\{f(x)\}[/tex]

[tex] \displaystyle \begin{aligned}\min\{f(x)\} &= f\left(\frac{24}{5}\right) \\ &= \sqrt{\left(\frac{24}{5}\right)^2+4}+\sqrt{\left(\frac{24}{5}\right)^2-24\left(\frac{24}{5}\right)+153} \\ &= \sqrt{\frac{576+100}{25}}+\sqrt{\frac{24}{5}\left(\frac{24-120}{5}\right)+153} \\ &= \sqrt{\frac{676}{25}}+\sqrt{\frac{-2304+3825}{25}} \\ &= \frac{26}{5}+\sqrt{\frac{1521}{25}} \\ &= \frac{26}{5}+\frac{39}{5} \\ &= \frac{65}{5} \\ &= 13 \end{aligned}\\[/tex]

Jawaban:

[tex] \displaystyle \boxed{\bold{\min\{f(x)\} = 13}}[/tex]

12. Maths Problem Terlampir

Jawab:

Penjelasan dengan langkah-langkah:

13. - The student ( Not Study) Maths+ The Student don't study Maths?​

Jawaban:

Does the student study maths?

Penjelasan:

Semoga membantu ^•^

maaf klw salah

Jawaban:

(?) are the students study maths?

Penjelasan:

buat introgatif tobe nya didepan ya, gw bingung ini yg positif nya gada tobe

14. 11th Quiz ☆Terlampir -,​

Jawaban:

Terlampir -,

by rai1194

Terlampiir caranya

832

15. QUIZ MATHS Terlampir

Jawaban:

130

Penjelasan dengan langkah-langkah:

27x + 28y + 29z = 363

karena 27 28 dan 29 adalah angka yang berdekatan maka 363 : (27+28+29) = 4,32

karena positif integer maka kita ambil bulatnya aja yaitu 4. Maka coba kombinasi angka 4 dan sekitarnya hingga dapat kombinasi

27 . 5 + 28 . 4 + 29 . 4 = 363

x = 5

y = 4

z = 4

maka

10x (5 + 4 + 4) = 100 x 13 = 1.300

16. QUIZ MATHS Terlampir

Penjelasan dengan langkah-langkah:

x^2+y^2=6

(x+y)^2 - 2xy = 6

(x+y)^2 - 2(2+3akar2-(x+y))=6

(x+y)^2+2(x+y)-(10+6akar2)=0

rumus ABC

[tex]x + y = \frac{ - b + - \sqrt{ {b }^{2} - 4ac} }{2a } \\ = \frac{ - 2 + - \sqrt{4 + 40 + 24 \sqrt{2} }}{2} \\ = \frac{ - 2 + - \sqrt{44 + 2 \sqrt{288} } }{2} \\ = - 1 + - (3 + \sqrt{2) } \\ = 2 + \sqrt{2} \\ atau \\ = - 4 - \sqrt{2} [/tex]

lx+y+1l= 3+akar 2

17. Quiz "Kombinasi dari :• Maths​

MathsM = 1a = 1t = 1h = 1s = 1------- +

C = 5! / 1! (5 - 1)!

= 5! / 1! (4)!

= 120 ÷ 24

= 5C

Maths

C = n! / r! ( n - r )!

C = 5! / 1! ( 5 - 1 )!

C = 5! / 4!

C = 120 / 24

C = 5

18. apa perbedaan 'she teaches maths' dan 'she teaching maths'​

Jawaban:

she teaches maths (Dia biasanya mengajar matematika)

She is teaching maths (Dia sedang mengajar matematika)

Jangan lupa di follow ya kak :)

Jawaban:

Kata tersebut memiliki perbedaan arti yaitu:

she teaches maths : Dia biasanya mengajar matematika

sementara → she teaching maths : Dia sedang belajar matematika.

Penjelasan:

.

.

Semoga Membantu~

19. 11th or 11st? ada yg bisa bantu?

Apabila 11 dijadikan tanggal, atau urutan. Maka, 11 menjadi 11th

Pembahasan

Hal tersebut dinamakan angka ordinal. Angka ordinal adalah angka dimana disusun secara berurutan dan terdapat tambahan seperti "st, nd, rd, atau th"

Ordinal Number [1-10]

1st => first. 6th => sixth

2nd => second. 7th => seventh

3rd => third. 8th => eighth

4th => fourth. 9th => ninth

5th => fifth. 10th => tenth

Dapat diketahui bahwa jika terdapat angka 1,2 atau 3 diakhir maka secara berurutan disebut first, second, third. Kecuali angka 11,12,13.

Apabila menemukan dua bilangan akhir 11,12,13 maka tetap menggunakan th dibelakang. Jadinya, eleventh dsb.

Detail Jawaban

Mapel : Bahasa Inggris

Kelas : 3 SD

Materi : Ordinal Number

Kata Kunci : 11th

Kode Kategorisasi : 3.5

#TingkatkanPrestasimu

#OptiTimCompetition

20. QUIZ MATHS terlampir

Jawab:

1003

Penjelasan dengan langkah-langkah:

[tex]\displaystyle f(x) = \frac{{9}^{x}}{{9}^{x}+3}[/tex]

Perhatikan bahwa :

[tex]\displaystyle f(x) + f(1 - x) = \frac{{9}^{x}}{{9}^{x}+3}+\frac{{9}^{1-x}}{{9}^{1-x}+3}\\= \frac{{9}^{x}({9}^{1-x}+3)+{9}^{1-x}({9}^{x}+3)}{({9}^{x}+3)({9}^{1-x}+3)}\\\\= \frac{{9}^{x}({9}^{1-x})+3\times{9}^{x}+9+3\times{9}^{1-x}}{9+3\times{9}^{x}+3\times{9}^{1-x}+9}\\=\frac{9+3\times{9}^{x}+9+3\times{9}^{1-x}}{9+3\times{9}^{x}+3\times{9}^{1-x}+9}\\\\=\frac{9+3\times{9}^{x}+9+3\times{9}^{1-x}}{9+3\times{9}^{x}+9+3\times{9}^{1-x}}\\\\=1[/tex]

maka :

[tex]\displaystyle f\left(\frac{1}{2007}\right)+f\left(1-\frac{1}{2007}\right)\\\\=f\left(\frac{1}{2007}\right)+f\left(\frac{2006}{2007}\right) = 1\\=f\left(\frac{2}{2007}\right)+f\left(\frac{2005}{2007}\right) = 1\\=f\left(\frac{3}{2007}\right)+f\left(\frac{2004}{2007}\right) = 1\\\\...\\...\\=f\left(\frac{1003}{2007}\right)+f\left(\frac{1004}{2007}\right) = 1[/tex]

[tex]\displaystyle \\-----------------------------------\:\:+\\\\f\left(\frac{1}{2007}\right)+f\left(\frac{2}{2007}\right)+f\left(\frac{3}{2007}\right)+f\left(\frac{4}{2007}\right)+...+f\left(\frac{2006}{2007}\right)=1\times1003\\\\=1003[/tex]