$nope$

bài kiểm tra toán thì bạn phải tự làm chứ

Bài em cần đâu em?

a: Xét ΔHAC vuông tại H và ΔABC vuông tại A có

\(\widehat{C}\) chung

Do đó: ΔHAC~ΔABC

b: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=15^2+20^2=625\)

=>BC=25

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}BH\cdot BC=BA^2\\AH\cdot BC=AB\cdot AC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH\cdot25=15^2=225\\AH\cdot25=15\cdot20=300\end{matrix}\right.\)

=>BH=9; AH=12

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

1. What will you do?
I will organize free tutoring classes, donate books and school supplies, and raise funds for students in need.

2. Why do you choose to do them?
Because education is very important, and I want to help students have better opportunities to learn and succeed.

3. Do you think the students in need will receive benefits? Why?
Yes, they will. They can improve their knowledge, have enough materials to study, and feel more confident about their future.

4. How can you make more people know about your voluntary activities?
I can share information on social media, invite friends to join, and cooperate with schools or local organizations to spread the message.

1 If you don't finish your homework, you can't go out with your friend

2 They are not sure how to operate the new system

3 I spent 4 hours reading the first chapter of the book

4 Tennis is not as dangerous as snowboarding

X

1 to go to school by bike when they were young

2 teaching her children to play the piano 4 years ago

3 I could cook as well as my mom

4 using Perfume Pagoda as the theme of the presentation

Giải thích zùm e đc ko ạ:((

🤔

🌚

a: Xét ΔADE có \(\frac{AB}{BD}=\frac{AC}{CE}\)

nên BC//DE
b: Ta có: \(\hat{DBM}=\hat{ABC}\) (hai góc đối đỉnh)

\(\hat{ECN}=\hat{ACB}\) (hai góc đối đỉnh)

mà \(\hat{ABC}=\hat{ACB}\) (ΔABC cân tại A)

nên \(\hat{DBM}=\hat{ECN}\)

Xét ΔDBM vuông tại M và ΔECN vuông tại N có

DB=EC

\(\hat{DBM}=\hat{ECN}\)

Do đó: ΔDBM=ΔECN

=>DM=EN

c: ΔDBM=ΔECN

=>\(\hat{BDN}=\hat{CEN}\)

AD=AB+BD

AE=AC+CE
mà AB=AC và BD=CE

nên AD=AE

Xét ΔADM và ΔAEN có

AD=AE

\(\hat{ADM}=\hat{AEN}\)

DM=EN

Do đó: ΔADM=ΔAEN

=>AM=AN

=>ΔAMN cân tại A

Bài 8:

Đặt CTTQ oxit kim loại hóa trị III là A2O3 (A là kim loại)

nH2SO4=0,3(mol)

mNaOH=24%. 50= 12(g) => nNaOH=0,3(mol)

PTHH: 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O

0,3________0,15(mol)

A2O3 +3 H2SO4 -> A2(SO4)3 +3 H2

0,05___0,15(mol)

=> M(A2O3)= 8/0,05=160(g/mol)

Mặt khác: M(A2O3)=2.M(A)+ 48(g/mol)

=>2.M(A)+48=160

<=>M(A)=56(g/mol)

-> Oxit cần tìm: Fe2O3

Bài 7:

mHCl= 547,5. 6%=32,85(g) => nHCl=0,9(mol)

Đặt: nZnO=a(mol); nFe2O3=b(mol) (a,b>0)

PTHH: ZnO +2 HCl -> ZnCl2+ H2O

a________2a_______a(mol)

Fe2O3 + 6 HCl -> 2 FeCl3 + 3 H2O

b_____6b____2b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}81a+160b=28,15\\2a+6b=0,9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\end{matrix}\right.\)

=> mFe2O3=0,1.160=16(g)

=>%mFe2O3=(16/28,15).100=56,838%

=>%mZnO= 43,162%