soal umpn kimia (UMPN 2019). Soal ada di gambar
1. soal umpn kimia (UMPN 2019). Soal ada di gambar
Jawaban:
52. C
53. C
Penjelasan:
52. BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
untuk mengetahui ini reaksi redoks atau bukan maka kita mencari biloks (bilangan oksidasi) ya dulu
BaCl2 = +2Na2SO4 = +4BaSO4 = +82NaCl = +1Nah setelah menentukan biloks kita tentukan reduksi dan oksidasi yang mana.
BaCl2 --> BaSO4 mengalami kenaikan biloks yg dari +2 ke +8 maka ini disebut oksidasi.
Na2SO4 --> 2NaCl mengalami penurunan biloks yg dari +4 ke +1 maka ini disebut reduksi.
Jadi, persamaan yang merupakan biloks adalah yang C
53. Al + H2SO4 --> Al2(SO4)3 + H2
nah ini anda hanya menentukan sisi kiri dan kanan agar sama.
Kiri: Kanan:
Al=1*2=2 Al= 2
H= 2*3=6 H= 2*3=6
S= 1*3=3 S= 3
O= 4*3=12 O= 12
Nah kita samakan itu koefisien dengan dikalikan.
Hasil
2Al + 3H2SO4 --> 1Al(SO4)3 + 3H2 C
Jawaban:
2. (B)
3. (C)
Penjelasan:
[tex]\fbox{2.}[/tex] [A] CaCO₃ (s) → CaO (s) + CO₂ (g)
Ruas kiri:
Ca = +2
C = +4
O = -2
Ruas kanan:
Ca = +2
C = +4
O = -2
Bukan reaksi redoks, karena tidak terjadi perubahan nilai biloks pada unsur-unsur yang bereaksi.
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
[B] Mg (s) + H₂SO₄ (aq) → MgSO₄ (aq) + H₂ (g)
Ruas kiri:
Mg = 0
H = +1
S = + 6
O = -2
Ruas kanan:
Mg = +2
H = 0
S = +6
O = -2
Merupakan reaksi redoks, karena terjadi perubahan biloks:
[tex]\checkmark\ Mg=0 \xrightarrow{\text{+}} +2\ \fbox{Oksidasi, kenaikan\ nilai\ biloks}[/tex]
[tex]\checkmark\ H=+1 \xrightarrow{\text{-}} 0\ \fbox{Reduksi, penurunan\ nilai\ biloks}[/tex]
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
[C] BaCl₂ (aq) + Na₂SO₄ (aq) → BaSO₄ (s) + 2NaCl (aq)
Ruas kiri:
Ba = +2
Cl = -1
Na = +1
S = +6
O = -2
Ruas kanan:
Ba = +2
Cl = -1
Na = +1
S = +6
O = -2
Bukan reaksi redoks, karena tidak terjadi perubahan biloks pada unsur-unsur yang bereaksi.
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
[D] Al₂O₃ (s) + 6HCl (aq) → 2AlCl₃ (aq) + 3H₂O (l)
Ruas kiri:
Al = +3
O = -2
H = +1
Cl = -1
Ruas kanan:
Al = +3
O = -2
H = +1
Cl = -1
Bukan reaksi redoks, karena tidak terjadi perubahan biloks pada unsur-unsur yang bereaksi.
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
[E] KOH (aq) + NH₄NO₃ (aq) → KNO₃ (aq) + NH₃ (aq) + H₂O (l)
Ruas kiri:
K = +1
O = -2
H = +1
N dalam NH₄ = -3
N dalam NO₃ = +5
Ruas kanan:
K = +1
O = -2
H = +1
N dalam NH₃ = -3
N dalam KNO₃ = +5
Bukan reaksi redoks, karena tidak terjadi perubahan biloks pada unsur-unsur yang bereaksi.
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
Jadi, pilihan yang sesuai adalah (B).
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
[tex]\fbox{3.}[/tex] Persamaan Reaksi: (belum setara)
Al (s) + H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + H₂ (g)
Ruas kiri:
Al = 1
H = 2
S = 1
O = 4
Ruas kanan:
Al = 2
H = 2
S = 3
O = 12
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
Penyetaraan Reaksi:
2Al (s) + 3H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + 3H₂ (g)
Ruas kiri:
Al = 2
H = 6
S = 3
O = 12
Ruas kanan:
Al = 2
H = 6
S = 3
O = 12
Berdasarkan reaksi tersebut, didapati koefisien reaksi secara berturut-turut adalah [tex]\fbox{2, 3, 1, 3}[/tex]
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
Jadi, pilihan yang sesuai adalah (C).
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
[tex]\fbox{Semoga\ Membantu}[/tex]
2. SOAL UMPN KIMIA POLBAN (UMPN 2019)
Jawaban:
122,5gram
Penjelasan:
pertama buat mula2, reaksi/terurai, sisa
kemudian cari mol KClO3 dan ubah ke gram
Jawaban:
(1) B. 122,5 gram
(2) B. +11,9 kkal
Penjelasan:
(1) [tex]\boxed{\star\ Persamaan\ Reaksi\ \star}[/tex]
[tex]KClO_{3}_{(aq)}+6\ HCl_{(aq)} \rightarrow KCl_{(aq)}+3\ H_{2}O_{(l)}+3\ Cl_{2}_{(g)}[/tex]
◈ Diketahui massa KCl yang terbentuk adalah sebanyak 74,5 gram.
◈ Massa molekul relatif KCl = 39 + 35,5 = 74,5 gram/mol
◈ Mol KCl = [tex]\frac{m}{Mr}[/tex] = [tex]\frac{74,5\ gram}{74,5\ gram/mol}[/tex] = 1 mol
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
[tex]\boxed{\boxed{Untuk\ mencari\ mol\ KClO_{3}, gunakan\ perbandingan\ koefisien\ pada\ reaksi.}}[/tex]
Mol KClO₃ = [tex]\frac{Koefisien\ KClO_3}{Koefisien\ KCl}\times mol\ KCl[/tex]
Mol KClO₃ = [tex]\frac{1}{1}\times 1\ mol[/tex]
Mol KClO₃ = [tex]1\ mol[/tex]
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
Mencari massa KClO₃ yang digunakan: (Mr KClO₃ = 122,5 gram/mol)
Mol KClO₃ = 1 mol
Mol KClO₃ = [tex]\frac{m}{Mr}[/tex]
Massa KClO₃ = [tex]mol \times Mr[/tex]
Massa KClO₃ = [tex]1\ mol \times 122,5\ gram/mol[/tex]
[tex]\boxed{Massa\ KClO_3 = 122,5\ gram}[/tex]
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
Jadi, massa kalium klorat yang diperlukan untuk menghasilkan 74,5 gram kalium klorida sebanyak dalam reaksi ini adalah sebanyak 122,5 gram.
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
(2) Diketahui perubahan entalpi untuk reaksi:
[tex]H_2O_{(l)} \rightarrow H_2_{(g)}+\frac{1}{2}\ O_2_{(g)}\ \ \Delta H = +68,3\ kkal\\H_2_{(g)}+\frac{1}{2}\ O_2_{(g)}\rightarrow H_2O_{(g)}\ \ \Delta H = -57,8\ kkal\\H_2O_{(l)} \rightarrow H_2O_{(s)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \Delta H = -1,4\ kkal[/tex]
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
Perubahan entalpi dari es menjadi uap air:
[tex]H_2O_{(l)} \rightarrow H_2_{(g)}+\frac{1}{2}\ O_2_{(g)}\ \ \Delta H = +68,3\ kkal\\H_2_{(g)}+\frac{1}{2}\ O_2_{(g)}\rightarrow H_2O_{(g)}\ \ \Delta H = -57,8\ kkal\\H_2O_{(s)} \rightarrow H_2O_{(l)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \Delta H = +1,4\ kkal[/tex]
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
Sehingga,
[tex]H_2O_{(s)} \rightarrow H_2O_{(g)} \ \ \ \ \ \ \ \ \Delta H =(68,3-57,8+1,4)kkal[/tex]
[tex]H_2O_{(s)} \rightarrow H_2O_{(g)} \ \ \ \ \ \ \ \ \Delta H = +11,9\ kkal[/tex]
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
Jadi, perubahan entalpi yang terjadi untuk mengubah es menjadi uap air adalah sebesar + 11,9 kkal.
[tex]\bigstar\ \bigstar\ \bigstar[/tex]
[tex]\fbox{Semoga\ Membantu}[/tex]
3. soal di gambar. (UMPN 2019)
Jawab:
Maaf tidak ada pada pilihan
Penjelasan dengan langkah-langkah:
Terlampir pada gambar....
4. Dapatnya 20 ( nomor 25). Tapi masih ragu tolong bantu dong. Soal Umpn 2015/2016 Tata niaga
f(x)=3x³-4x²+5x+2
f'(x)=9x²-8x+5
f"(x)=18x-8
f"(2)=18(2)-8
f"(2)=36-8
f"(2)=28 [D]
-SkyeDz-
5. adalah... tolong dong jawab soal prediksi UMPN
Mapel : Matematika
Kelas : X SMA
Bab : Trigonometri
Pembahasan :
Ada pada gambar...
Jawaban tidak ada di Opsi.
6. soal UMPN Polban 2019
Jawaban:
27. B
28. B
Penjelasan dengan langkah-langkah:
Ingat ingat lagi sifat-sifat pangkat dan rumus integral tak tentu, semangatt
27.)[tex] \int \frac{1}{(x-1)^{\frac{3}{2}}} \: dx = ....?\\[/tex]
Penyelesaian:
misal:
[tex] u = x-1 \\ du = dx\\ [/tex]
[tex] \begin{aligned} \int \frac{1}{(x-1)^{\frac{3}{2}}} \: dx &= \int u^{-\frac{3}{2}} \: du \\ &= \frac{1}{-\frac{3}{2}+1}u^{-\frac{3}{2}+1}+C \\ &= \frac{1}{-\frac{1}{2}} u^{-\frac{1}{2}}+C \\ &= -\frac{2}{\sqrt{u}}+C \\ &= -\frac{2}{\sqrt{x-1}}+C\end{aligned}[/tex]
Jawaban:
[tex]\bold{B. \; -\frac{2}{\sqrt{x-1}}+C} [/tex]
28.)[tex] \int_{2}^{4} \frac{t^2-1}{t-1} \: dt\\[/tex]
Penyelesaian:
[tex]\begin{aligned}\int_{2}^{4} \frac{t^2-1}{t-1} \: dt &= \int_{2}^{4} \frac{(t+1)(t-1)}{t-1} \: dt \\ &= \int_{2}^{4} t+1 \: dt \\ &= \frac{1}{2}t^2+t \Big|_{2}^{4} \\ &= \left(\frac{1}{2}(4)^2+(4) \right) - \left(\frac{1}{2}(2)^2+(2) \right) \\ &= (12)-(4) \\ &= 8 \end{aligned}\\[/tex]
Jawaban:
[tex]\bold{B. \; 8}[/tex]
7. Apakah materi soal umpn pada gelombang 1 dan gelombang 2 sama?
Setauku sama, cuma beda waktunya aja..
8. Tolong bantuin ,soal umpn itu fisika
v pendengar menjauhi sumber bernilai negatif dan v sumber mendekati pendengar bernilai negatif juga.
9. Bang soal bahasa Inggriswrite download 10 example sentence of connected speech
Jawaban:
payphone you dring koko
10. download soal olimpiade bahasa indonesia sma dan pembahasannya
Jawaban:
maksudnya ga ngerti
Penjelasan:
maaf ya
11. soal umpn 2019pernyataan berikut yang tidak sesuai dengan isi tabel di atas adalah...
Jawaban:
A. Jumlah penjualan aspal tertinggi pada bulan Juni
C. Drum yang terjual pada bulan April berjumlah 260 buah.
SEMOGA bermanfaat
Penjelasan:
yang tidak sesuai:
(A),(B),(C)
12. download software kamus suara bahasa inggris
Kelas: SMP
Pelajaran: English
Kategori: Useful resource
Pembahasan:
For the dictionary usage in the smartphone, try to download, Wiktionary application, in Playstore. It is very complete, including the expressions, idioms and phrasal verbs in English as well as equipped with recorded sound of pronouncing the vocabulary. What's more, this English-English dictionary can work offline too.
13. soal ada di gambar. tolong jawab dua soal itu ya (UMPN 2019)
23.)
[tex]f(x)=2\sqrt{x}(3x^2-4)\\f'(1)=\dots?[/tex]
Penyelesaian:
misal:
[tex]\begin{array}{ll}u = 2\sqrt{x} & v = 3x^2-4\\u' = \frac{1}{2}\cdot2x^{\frac{1}{2}-1} & v'= 6x^{2-1}-0 \\u' = \frac{1}{\sqrt{x}} & v' = 6x\end{array}{cc}[/tex]
maka
[tex]\begin{aligned}f'(x) &= u'v+uv'\\f'(x) &= \left(\frac{1}{\sqrt{x}}\right)(3x^2-4)+(2\sqrt{x})(6x)\\\text{Jadi,}\\\end{aligned}\\\begin{aligned}f'(1) &= \left(\frac{1}{\sqrt{(1)}}\right)(3(1)^2-4)+(2\sqrt{(1)})(6(1))\\&= (3-4)+(2)(6)\\&= -1+12\\&= 11\end{aligned}[/tex]
Jawaban:
[tex]\bold{D. \; 11}[/tex]
24.)[tex]f(x) = \frac{8}{\sqrt{x}}\\ f''(2) = \cdots?[/tex]
Penyelesaian:
[tex]\begin{aligned} f(x) &= 8x^{-\frac{1}{2}}\\ f'(x) &= -\frac{1}{2}\cdot8 x^{-\frac{1}{2}-1}\\ f''(x) &= -\frac{3}{2}\cdot(-4)x^{-\frac{3}{2}-1}\\ &= \frac{6}{\sqrt{x^5}}\\ f''(2) &= \frac{6}{\sqrt{(2)^5}}\\ &= \frac{6}{4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\ &= \frac{3}{2\cdot2}\sqrt{2}\\ &= \frac{3}{4}\sqrt{2} \end{aligned}[/tex]
Jawaban:
[tex]{\bold{C. \; \frac{3}{4}\sqrt{2}}}[/tex]
14. soal di gambar. (UMPN 2019)
Jawab: D
Penjelasan dengan langkah-langkah:
Terlampir pada gambar...
nb: Gunakan aturan cos
Semoga Membantu kak...
Jawab:
D. [tex]\frac{2q^{2}-p^{2}}{2q^{2}}[/tex]
Penjelasan dengan langkah-langkah:
∠AOB = θ
AB = p
OA = OB = q
AB² = OA² + OB² - 2.OA.OB.cosθ
p² = q² + q² - 2.q.q.cosθ
p² = 2q² - 2q².cosθ
2q².cosθ = 2q² - p²
cosθ = (2q² - p²) / 2q²15. apa bahasa bali download???
tetep download setau saya pas tinggal di sana dulu tetep download lah. ga diubah ubah lagi
16. soal ada di gambar. (UMPN 2019)
sin a = de/mi
cos a = sa/mi
tan a = de/sa
kuadran I = sin cos tan bernilai +
kuadran II = hanya sin bernilai +
kuadran III = hanya tan bernilai +
kuadran IV = hanya cos bernilai +
cos b = -√5/4
Sisi samping = √5 dan Sisi miring=4
Sisi depan² = sisi miring² - Sisi samping²
Sisi depan² = 4² - (√5)² = 16 - 5 = 11
Sisi depan = √11
tan b = de/sa = √11/√5 x √5/√5
tan b = √55/5
jawabannya D
smoga membantu
17. Soal UMPN 2017 tolong bantu ya...
semoga bisa membantu belajar
kalo ada yang kurang jelas ditanyakan yaaa^^
aku juga peserta umpn
kamu ikut umpn gelombang kedua atau gimana?^^
18. Tolong soal umpn matematika 2017
f(x) = (2x^2 + 3)/(x - 6) ==> f(0) = (2.0^2 + 3)/(0 - 6) = 3/-6 = -1/2
f(x) = u/v
u = 2x^2 + 3 => u' = 4x
v = x - 6 => v' = 1
f'(x) = (u' v - v' u)/v^2
f'(x) = (4x (x - 6) - 1(2x^2 + 3)) / (x - 6)^2
f'(0) = (4(0) (0 - 6) - (2(0)^2 + 3)) / (0 - 6)^2
f'(0) = (0 - 3) / (-6)^2
f'(0) = -3/36 = -1/12
f(0) + f'(0) = -1/2 + (-1/12) = -6/12 - 1/12 = -7/12Gitu ka mungkin aalah soalnya wkwk
19. Soal Umpn 2015/2016 tata niaga
8. Jumlah mata dadu minimal 10
berarti kombinasi jumlah yang mungkin adalah 4+6, 6+4, 5+5, 5+6, 6+5, 6+6
Banyaknya sebanyak 6
maka peluangnya = 6/36 = 1/6
20. dimana kita bisa download rpp bahasa inggris kelas 8 terlengkap?
Jawaban:
Maybe bisa tanya guru. Atau coba cari di google gitu "RPP kls8 terlengkap"?
kan hope it's useful
[tex]andynakiarra28[/tex]
21. Soal UMPN 2018 bidang rekayasa
√x +1/√x =3
Kuadrat kan kedua ruas
x+2+1/x =9
x+1/x =7
x¹+x^-1 =7
22. Soal UMPN KIMIA POLBAN (UMPN 2019) jawab no 58 dan 59 ya. terimakasih
no 58)
4NH3 + 5O2 ---> 4NO + 6H2O
berarti
* laju O2= Koef. O2/koef NH3 x laju NH3 = 5/4 x 0,24 = 0,30
* laju pmbntkn H2O = koef H2O/ koef NH3 x laju NH3 = 6/4 x 0,24 = 0,36
no 59)
berarti
M1.V1 = M2.V2
10.0,1 = M2.(10+90)
M2 = 1/100 = 1.10^-2 Mol
pH awal (HCl 10 ml) =
-log[H+] = -log[M1]
pH = -log10^-1
pH = 1
pH akhir (HCl 10+90 ml = 100 ml) =
-log[H+] = -log[M2]
pH = -log10^-2
pH = 2
pH awal = 1
pH akhir = 2
Jawaban:
58. C 0,30 dan 0,36
semoga membantu
23. download pidato bahasa inggris yan banyak kosa kata bahasa inggris
contoh pidato(nama sekolah hanya contoh)
Education in Indonesian
The honorable ones the principle of EDUCATION Kristen Immanuel 2, all teachers and administration staffs of EDUCATION Kristen Immanuel 2 and all my friends from Kindergarden, and one until six grade.
Before I begin to give a speech, I would like to invite you all to thank Allah the Almighty, Who has given us Mercy and Blessing, so we can meet together in this blessing place. And also let’s deliver sholawat and salam to our Jesus Christ, Who has brought us from the darkness to the brightness, so we are always in the right path.
Ladies and gentlemen
It be a great honor for me, to stand right here in front of you all to give a speech about education under the title "Education in Indonesia" in this National Education Day or HARDIKNAS.
Indonesian is a great country. It has so many natural resources that can be used. It has so many beautiful palaces that can mesmerize people. It is so beautiful to live here peacefully with different cultures, languages, and religions. However, why are there still many people who live under poverty in this country? The only answer to this question is because our human resources are still low in the term of education.
If you have ever heard about the news about our international education rank, you would be surprised. As quoted form Education for All (EFA) Global Monitoring Report 2011 which is published by UNESCO and launched in New York on Sunday, 1/3/2011, our international education rank is at 69th place from 129 surveyed countries all over the world. This rank is lower than in 2010.
It opens a question that runs in my head, what happen with our education? Indonesia is a great country with many talented young generations. It was undoubted again that there are so many Indonesian students who win the international education Olympiads such as mathematics, chemical, physics and other. But why our international education rank is still low. So what happen with with our education?
Ladies and gentlemen
There must be a reason why our human resources and our global rank in education are still low. One of the reasons is that our education is expensive. For middle up to high class people the cost of our education is nothing. But for those who live in lower economic class, the cost of our education is beyond their reach. Many of them even abandon their duty to go to school just because they have to work to fulfill their needs.
Our government has made a policy recently to overcome the problem in our education. The policy is increasing national educational budget up to 20% from APBN. This budget later on is used to make education is reachable for low economic level society; to improve teachers’ quality with a program which well known as “sertifikasi”; and to provide school facilities such as library, classroom, laboratory, and sport hall.
Ladies and gentlemen
We hope the policy that the government made will make our educational quality better. Of course as a student, we don’t just sit down and relax without doing something. Good or bad the quality of education of our country also depend us. If we want to change our bad educational paradigm, let’s begin with improving our achievement with studying hard for the sake of our education.
I hope this speech that I deliver today will be useful for you all. Thank you so much for your attention. The last I say:good morning.
24. Tolong Kakakakakkakakak..,...... Soal Umpn
Terlampir caranya.. semoga membantu :)
25. Soal UMPN KIMIA POLBAN (UMPN 2019) jawab no 58 dan 59 ya. terimakasih
Jawaban:
maaf cuma bisa bantu jawab no. 59 jawabannya A. 1 menjadi 2
semoga membantu:)
26. pengaturan pelaksanaan umpn dilaksanakanmohon bantuannya:'(
Jawaban:
pertanyaannya apa?
Penjelasan:
diperjelas lagi pertanyaannya
27. bahasa inggrisnya ini apa? jadi gini men temen aku download tu nggak ada stok foto lagi?
Jawaban nya adalah this ini adalah THIS
semoga membantu
28. Soal umpn 2017 matematika no 28, ,29,30
28) (p ... q) (x) = (p)
...... (q ... p) (y) ... (q)
px + qy = p
qx + py = q
---------------- +
px + qx + qy + py = p + q
(p + q)x + (p + q)y = (p + q) ======> kedua ruas dibagi (p + q)
x + y = 1
29) AB = I
B = A^-1 . I
B = A^-1 = 1/-2 (1 ... -1) = (-1/2 ... 1/2)
......................... (-2 .. 0) .. (1 .......... 0)
30) (2 ... a) (x) = (5)
....... (b .. 6) (y) .. (7)
2x + ay = 5 => m1 = -x/y = -2/a
bx + 6y = 7 => m2 = -x/y = -b/6
Karena sejajar maka
m1 = m2
-2/a = -b/6
ab = 12
29. Soal UMPN 2019, Kasih caranya tolong
Jawab:
Fungsi
komposisi dan invers
i) (fog)(x) = f { g(x)}
ii) (fog)⁻¹ (x) = (g⁻¹ o f⁻¹)(x) = g⁻¹ { f⁻¹(x)}
Penjelasan dengan langkah-langkah:
f(x)= (2x - 5) / (x)
f⁻¹ (x) = ( -5)/ (x - 2)
.
g(x) = 2x - 3
g⁻¹ (x) = (x + 3)/2
.
(fog)⁻¹ (x) = (g⁻¹ o f⁻¹)(x) = g⁻¹ { f⁻¹(x)}
(fog)⁻¹ (x) = g⁻ ¹ { -5/(x - 2) }
(fog)⁻¹ (x) = { -5/(x - 2) + 3 } / (2)
(fog)⁻¹ (x) = { -5 + 3 (x - 2)} / (2)(x - 2)
(fog)⁻¹ (x) = (-5 + 3x - 6 )/ (2x - 4)
(fog)⁻¹ (x) = (3x - 11)/ (2x - 4)
..
cara lain
i) cari (fog)(x)= f { g(x)}
ii) inverskan
dilampiran
30. Apakah anak IPS bisa masuk teknik di PTN atau POLITEKNIK lewat SBMPTN atau UMPN
Jawaban:
bisa jika ada niat usaha serta doa
31. bagaimana cara mengnyekesaikan ujian umpn
dengan cara jujur, selalu berdoa, berusaha dan tawakal
32. Mohon di bantu menjawab pertanyaan umpn
[tex]2( \sqrt{5 } - 1) - \frac{8}{ \sqrt{5} - 1} \\ = 2 \sqrt{5} - 2 - \frac{8}{ \sqrt{5} - 1} \times \frac{ \sqrt{5} + 1 }{ \sqrt{5} + 1} \\ = 2 \sqrt{5} - 2 - \frac{8( \sqrt{5} + 1)}{5 - 1} \\ = 2 \sqrt{5} - 2 - 2( \sqrt{5} + 1) \\ = 2 \sqrt{5} - 2 - 2 \sqrt{5} - 2 = - 4[/tex]
Jawaban: D.
[tex]\displaystyle 2(\sqrt5-1)-\frac{8}{\sqrt5-1}=2(\sqrt5-1)-\frac{8}{\sqrt5-1}\cdot\frac{\sqrt5+1}{\sqrt5+1}\\2(\sqrt5-1)-\frac{8}{\sqrt5-1}=2(\sqrt5-1)-\frac{8(\sqrt5+1)}{5-1}\\2(\sqrt5-1)-\frac{8}{\sqrt5-1}=2(\sqrt5-1)-2(\sqrt5+1)\\2(\sqrt5-1)-\frac{8}{\sqrt5-1}=2\sqrt5-2-2\sqrt5-2\\\boxed{\boxed{2(\sqrt5-1)-\frac{8}{\sqrt5-1}=-4}}[/tex]
33. bagaimana saya bisa mendownload soal bahasa indonesia smk dengan mudah
kamu bisa mencarinya di link ini
http://shellypratiwiningsih.blogspot.co.id/2014/09/kumpulan-soal-bahasa-indonesia-smk-dan.html?m=1
semoga bermanfaat
34. aku download brainly for iOS tapi kenapa jadi bahasa inggris yaa?
coba aja di settingan ato ga di unistall trs install lg (?) maaf klo ga ngebantu :")
35. Tugas b.inggris 1).bagaimana cara membuat bbm dalam bahasa inggris 2).bagaimana cara mendownload aplikasi dalam bahasa inggris
coba rubah pengaturan bahasa.
36. Tolong gan 6-8 soal Umpn 2018/2019 tata niaga
saya jawab soal nomor 7&8
7. C=A-B
C = [tex] \left[\begin{array}{ccc}5-(-2)&2-3\\3-2&1-(-1)\end{array}\right] [/tex]
C = [tex] \left[\begin{array}{ccc}7&-1\\1&2\end{array}\right] [/tex]
[tex] C^{-1} = \frac{1}{det C} [adj C] [/tex]
det C = ad - bc
det C = (7×2)-(-1×1)
det C = 14-(-1) = 14 + 1 = 15
adj C = [tex] \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right] [/tex]
adj C = [tex] \left[\begin{array}{ccc}2&1\\-1&7\end{array}\right] [/tex]
[tex] C^{-1} = \frac{1}{15} \left[\begin{array}{ccc}2&1\\-1&7\end{array}\right] [/tex]
8. [tex] \frac{1}{3}\sqrt{\frac{27}{12}} - \sqrt[3]{\frac{8}{27}} = [/tex]
[tex] \frac{1}{3} \sqrt{\frac{9 . 3}{4 . 3}} - \sqrt[3]{\frac{2.2.2}{3.3.3}} = [/tex]
[tex] \frac{1}{3}. \frac{3}{2} - \frac{2}{3} = [/tex]
[tex] \frac{3}{6} - \frac{2}{3} = \frac{3-4}{6} = -\frac{1}{6} [/tex]
semoga membantu :D
37. soal di gambar. (UMPN 2019)
Jawaban:
C
Penjelasan dengan langkah-langkah:
[tex]\frac{5}{x - 7} > \frac{7}{x + 5} \\ \frac{5}{x - 7} - \frac{7}{x + 5} > 0 \\ \frac{5(x + 5) - 7(x - 7)}{(x - 7)(x + 5)} > 0 \\ \frac{74 - 2x}{(x - 7)(x + 5)} > 0
[/tex]
Pembuat nol pembilang :
[tex]74 - 2x = 0 \\ x = 37[/tex]
pembuat nol penyebut
[tex]x = 7 \: atau \: x = - 5[/tex]
buat garis bilangan :
--------(-5)----------------(7)----------------(37)
uji angka misal subtitusi x = 0 maka
interval -5 sampai 7 menjadi negatif maka garis bilangannya menjadi :
POS. NEG. POS. NEG
--------(-5)----------------(7)-------------(37)--------
karena yang diinginkan adalah positif, maka jawabannya adalah
[tex]x < - 5 \: atau \: 7 < x < 37[/tex]
38. soal terlampir. umpn kimia 2019
berarti
jumlah mol Cu = 40/64 = 0,625
jumlah mol S = 4/32 = 0,125
2Cu. + S ---> Cu2S
awal 0,6125 0,125. -
reaksi 0,25. 0,125. 0,25
akhir. 0,375. - 0,125
sisa 0,375 Cu yang gak bereaksi
berat zat yg gak bereaksi =
0,375 x 64 = 24 gram
Jawaban:
(B.) 24 gram
Penjelasan:
Pertama, cari jumlah mol masing-masing zat yang bereaksi.
Mol Cu = [tex]\frac{m}{Ar}[/tex] = [tex]\frac{40\ gram}{64\ sma}[/tex] = 0,625 mol
Mol S = [tex]\frac{m}{Ar}[/tex] = [tex]\frac{4\ gram}{32\ sma}[/tex] = 0,125 mol
⟡ ⟡ ⟡
Kemudian, kita reaksikan kedua zat tersebut.
[tex]\boxed{\bold{Persamaan\ Reaksi}}[/tex]
[tex]\ \ \ \ \ \ 2Cu_{(s)}\ +\ \ \ S_{(s)}\ \ \ \rightarrow\ \ Cu_2S_{(s)}\\m\ \ .625\ mol\ \ \ .125\ mol\ \ \ \ \ \ \ \ \ - \\r\ \ \ .250\ mol\ \ \ .125\ mol\ \ \ \ \ \ .125\ mol\\s\ \ \ .375\ mol\ \ \ \ \ \ \ -\ \ \ \ \ \ \ \ \ \ .125\ mol[/tex]
⟡ ⟡ ⟡
Di akhir reaksi masih terdapat sisa mol Cu sebanyak 0,375 mol. Sehingga massa zat yang tidak bereaksi tersebut adalah:
Massa Cu = mol Cu sisa × Ar
Massa Cu = 0,375 mol × 64 sma
[tex]\boxed{Massa\ Cu=24\ gram}[/tex]
⟡ ⟡ ⟡
Jadi, massa zat tembaga yang tidak bereaksi dengan belerang dalam reaksi tersebut adalah sebanyak 24 gram.
⟡ ⟡ ⟡
[tex]\boxed{\star\ Semoga\ Membantu\ \star}[/tex]
39. SOAL UMPN FISIKA POLBAN 2019 Tolong jawab nomor 32 dan 33 ya. Terimakasih
Jawaban:no 33
Penjelasan:
Diket: m1= 70
m2 = 30
g=10m/s
A= m1-m2/m1+m2.g
= 70-30/70+30.10
= 40/100.10
= 4 m/s2
40. Mohon di bantu yahhh..... Soal umpn...2 soal....
Uraian dalam lampiaran
semoga membantu
