Camon ạ
Giúp mik câu 2 vs ạ. Camon ạ
Bài 1:
a) Xét ΔMNQ và ΔENQ có
NM=NE(gt)
\(\widehat{MNQ}=\widehat{ENQ}\)
NQ chung
Do đó: ΔMNQ=ΔENQ(c-g-c)
Suy ra: QM=QE(hai cạnh tương ứng)
Bài 1:
b) Ta có: ΔQMN=ΔQEN(cmt)
nên \(\widehat{QMN}=\widehat{QEN}\)(hai góc tương ứng)
mà \(\widehat{QMN}=90^0\)(gt)
nên \(\widehat{QEN}=90^0\)
Giúp mik bài này với ạ . CAMON trc ạ 🥰
1 He is told to take a long rest
2 This plant isn't watered everyday
3 Beer is drunk all over the world
4 Uniform is worn to school by students
5 Reports are sent to the manager by the secretary in the afternoon
6 How many languages are spoken in Canada?
Bài 2
1 I was brought this dish by the waiter
2 I wasn't showed the special cameras
3 Her ticket was showed to her friends
4 The broken cup was hidden in the drawer by Tom
5 The fridge wasmoved into the living room by his uncle
mng ơi mng giúp em vs ạ em camon ạ
a: Ta có: \(\widehat{ABC}=\widehat{ACB}\)(ΔABC cân tại A)
\(\widehat{ACB}=\widehat{ECN}\)(hai góc đối đỉnh)
Do đó: \(\widehat{ABC}=\widehat{ECN}\)
Xét ΔMBD vuông tại D và ΔNCE vuông tại E có
BD=CE
\(\widehat{MBD}=\widehat{NCE}\)
Do đó: ΔMBD=ΔNCE
=>DM=EN
b: Ta có: DM\(\perp\)BC
EN\(\perp\)BC
Do đó: DM//EN
Xét ΔIDM vuông tại D và ΔIEN vuông tại E có
MD=EN
\(\widehat{MDI}=\widehat{ENC}\)(hai góc so le trong, DM//EN)
Do đó: ΔIDM=ΔIEN
=>IM=IN
=>I là trung điểm của MN
giúm mình bài này v ạ so sánh ạ mình camon
38.
$\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}< \frac{1}{\sqrt{2005}+\sqrt{2004}}=\frac{2005-2004}{\sqrt{2005}+\sqrt{2004}}=\sqrt{2005}-\sqrt{2004}$
39.
$\sqrt{1998}+\sqrt{2000}-2\sqrt{1999}=(\sqrt{2000}-\sqrt{1999})-(\sqrt{1999}-\sqrt{1998})$
$=\frac{1}{\sqrt{2000}+\sqrt{1999}}-\frac{1}{\sqrt{1999}+\sqrt{1998}}$
$< 0$
$\Rightarrow \sqrt{1998}+\sqrt{2000}<2\sqrt{1999}$
Bài 40 bạn làm tương tự câu 38.
giúp mình vs ạ,camon
d: Ta có: \(12x^2+7x-12\)
\(=12x^2+16x-9x-12\)
\(=4x\left(3x+4\right)-3\left(3x+4\right)\)
\(=\left(3x+4\right)\left(4x-3\right)\)
e: Ta có: \(15x^2+7x-2\)
\(=15x^2+10x-3x-2\)
\(=\left(3x+2\right)\left(5x-1\right)\)
giúp mik ạ, camon nhìu
bài 5 gấp ạ camon
\(\left\{{}\begin{matrix}P+N+E=40\\P=E\\\left(P+E\right)-N=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2P+N=40\\2P-N=12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}N=14\\P=E=Z=13\end{matrix}\right.\)
=> Điện tích hạt nhân: Z+= 13+
Sơ đồ cấu tạo nguyên tử X:
mng ơi giúp mình vs ạ .mình cần gấp .mk camon ạ
1............to work by car everyday
2 ........Khanh does morning exercise, he won't keep fit
3......has a small garden
4 .....has to go to school on time
5 ......she doesn't go to bed early, she will be late for class
6 .......go to the movie tonight
\(\text{⩷!⨙┏⇴⨗⨜}\)
Giúp em với ạ e camon
Ex6
1 was
2 had been
3 found
4 had talked
5 had spoken
6 snowed
7 were
8 painted
9 had
10 had known
Ex7
2 had got
3 had hurt
4 were
5 were
6 was
7 wanted
8 ?
9 stopped
10 had eaten
Giúp em vs ạ ! Em camon.
\(d,\left(2x-1\right)\left(x+3\right)+2x\left(2-x\right)=10-8\left(x+4\right)\\ \Leftrightarrow2x^2+6x-x-3+4x-2x^2=10-8x-32\\ \Leftrightarrow17x=19\Leftrightarrow x=\dfrac{19}{17}\)
vậy phương trình đã cho có nhiệm \(x=\dfrac{19}{17}\)
Ta có: \(\left(2x-1\right)\left(x+3\right)+2x\left(2-x\right)=10-8\left(x+4\right)\)
\(\Leftrightarrow2x^2+6x-x-3+4x-2x^2=-8x-22\)
\(\Leftrightarrow18x=-19\)
hay \(x=-\dfrac{19}{18}\)