Giúp mình câu 29
Giúp mình câu 29 làm theo kiểu mảng giúp mình
29:
uses crt;
var n,i,j,kt:integer;
a:array[1..100]of integer;
begin
clrscr;
readln(n);
for i:=1 to n do
readln(a[i]);
for i:=1 to n do
if a[i]>1 then
begin
kt:=0;
for j:=2 to trunc(sqrt(a[i])) do
if a[i] mod j=0 then kt:=1;
if kt=0 then write(a[i]:4);
end;
readln;
end.
GIÚP MÌNH CÂU 29 --> CÂU 40
29 better
30 consumption
31 qualified
32 environmentalists
33 destroyed
34 abruptedly
VIII
35 Tet is the best time when family member from all over the country come back home
36 The parents felt happy when they were informed about their children's examination results
IX
1 Because of performing excellently at AFC tournament, everybody admired U23 VN
2 I have not heard that news from my close friend for 2 days
3 Robinson Crusoe novel, which was bought by my father last week, is interesting
4 Our teacher advised us to wear medical face mask when going out
Giúp mình câu 29 nhé!
\(f'\left(x\right)=\dfrac{-3}{\left(x-1\right)^2}\) ; \(f''\left(x\right)=\dfrac{6}{\left(x-1\right)^3}\)
\(f'\left(x\right)+f''\left(x\right)=0\Leftrightarrow\dfrac{-3}{\left(x-1\right)^2}+\dfrac{6}{\left(x-1\right)^3}=0\) (\(x\ne1\))
\(\Leftrightarrow-3\left(x-1\right)+6=0\Rightarrow x=3\)
Giúp mình câu 28 29
28)
$n_{CO_2} = \dfrac{6,72}{22,4} = 0,3(mol)$
$n_{H_2O} = \dfrac{9,9}{18} = 0,55(mol)$
Ta thấy : $n_{H_2O} > n_{CO_2} \to$ Ancol no
$\Rightarrow n_{O\ trong\ X} = n_{OH} = n_{H_2O} - n_{CO_2} = 0,25(mol)$
$\Rightarrow m_X = m_C + m_H + m_O = 0,3.12 + 0,55.2 + 0,25.16 = 8,7(gam)$
Khi ete hóa : $n_{H_2O} = \dfrac{1}{2}n_{OH} = 0,125(mol)$
$\Rightarrow m_{ete} = m_X - m_{H_2O} = 8,7 - 0,125.18 = 6,45(gam)$
Đáp án A
29)
$n_{CO_2} = 0,3(mol) ; n_{O_2} = 0,35(mol)$
Bảo toàn C :
$n_C = n_{CO_2} = 0,3(mol)$
Số nguyên tử C = $\dfrac{n_C}{n_X} = \dfrac{0,3}{0,1} = 3$
Loại : đáp án A
X là ancol no $\to n_{H_2O} - n_{CO_2} = n_X \Rightarrow n_{H_2O} = 0,4(mol)$
$\Rightarrow n_H = 2n_{H_2O} = 0,8(mol)$
Số nguyên tử H $= \dfrac{n_H}{n_X} = 8$
Bảo toàn nguyên tố C :
$n_{OH} + 2n_{O_2} = 2n_{CO_2} + n_{H_2O} \to n_{OH} = 0,2$
Số nhóm $OH = \dfrac{n_{OH}}{n_X} = 2$
Vậy CT của X là $C_3H_6(OH)_2$
Giúp mình câu 29 30
Giúp mình câu 29 30
Giúp mình câu 29 30
Giúp mình câu 29 30
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GIÚp mình câu 29 với ạ
Giúp mình 29 câu này với
2 The man about whom we are talking taught us English
3 The matter to which you should pay attention is of great important
4 The woman to whom you gave place on the bus lives next door to me
5 The school at which we are studying is too large
6 THe girl with whom I went on holiday is so cute
7 The city in which my grandmother live seems to be quiet
8 It is the shop in which my sister works
9 They are the interesting books for which I have been looking
10 The picnic to which I went yesterday morning was really exciting
14 The film The eigh-year-old bride about which Mai talked is very too long
15 Ba Be National park about which my friends talking is very beautiful
16 The college at which we used to study was large with a large ground
17 Mrs Hoa, about whom the students are talking, is very lovely and gentle
18 The villa in which they lived is not only beautiful but also very modern
19 Mr Vu, with whom I made friend, is a handsome and strong man
20 The novel about which I told you is about the adventure of a cricket
21 The widow with whom my brother fell in love is very pretty
22 Mrs Hoa and Mrs Quyen, with whom I work, teach English very well
23 Ha Noi, in which we are going to live, is a beautiful city
24 The crash about which I and my friends are talking is really terrible
25 Bach Ma National Park, in which many kind of rare animal are living, is very beautiful
26 The banquat to which we went yesterday was very delicious
27 The banchelor to whom I talked last night was very kind
28 The flat in which we used to live was large with a green garden
29 Nam, with whom I learn in the same class, is a good student