Giúp mnh câu c,d nhe
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TC
21 tháng 12 2022
Lực đẩy Ac si met là
`F_A = P-P_n = 2,13-1,83=0,3(N)`
Do vật chìm hoàn toàn nên ta có
thể tích của vật là
`V_(vật)= F_A/d_n=(0,3)/10000=3*10^(-5)(m^3)=30(cm^3)`
L
16 tháng 1 2022
1.I_______ to Ho Chi Mnh City twice.
A. have gone B. go C. went D. goes
2. We ___________the beaches so far.
A. visit B. have visited C. visited D. are visiting
3. He ___________her before.
A. never met B. has never met C. never meets D. never meet
4. I __________ this picture many times.
A. draw B. have drawn C. drew D. am drawing
5. She ___________ milk tea before.
A. never drinks B. never drank C. has never drunk D. never drink
L
16 tháng 1 2022
II. VERB FORM (05 câu)
1.I_______ to Ho Chi Mnh City twice.
A. have gone B. go C. went D. goes
2. We ___________the beaches so far.
A. visit B. have visited C. visited D. are visiting
3. He ___________her before.
A. never met B. has never met C. never meets D. never meet
4. I __________ this picture many times.
A. draw B. have drawn C. drew D. am drawing
5. She ___________ milk tea before.
A. never drinks B. never drank C. has never drunk D. never drink
16 tháng 1 2022
Mk lm hz chậm
II. VERB FORM (05 câu)
1.I_______ to Ho Chi Mnh City twice.
A. have gone B. go C. went D. goes
2. We ___________the beaches so far.
A. visit B. have visited C. visited D. are visiting
3. He ___________her before.
A. never met B. has never met C. never meets D. never meet
4. I __________ this picture many times.
A. draw B. have drawn C. drew D. am drawing
5. She ___________ milk tea before.
A. never drinks B. never drank C. has never drunk D. never drink
N
2
26 tháng 12 2023
a: \(\dfrac{1}{3}:x=2\dfrac{2}{3}:\left(-0,3\right)\)
=>\(\dfrac{1}{3}:x=\dfrac{8}{3}:\dfrac{-3}{10}=-\dfrac{80}{9}\)
=>\(x=-\dfrac{1}{3}:\dfrac{80}{9}=-\dfrac{1}{3}\cdot\dfrac{9}{80}=-\dfrac{3}{80}\)
b: \(3^{2x}-2\cdot3^5=3^5\)
=>\(3^{2x}=3^5+2\cdot3^5=3\cdot3^5=3^6\)
=>2x=6
=>x=3
c: \(2x-\sqrt{1,69}=\sqrt{1,21}\)
=>2x-1,3=1,1
=>2x=1,3+1,1=2,4
=>x=2,4/2=1,2
d: \(\left|x+\dfrac{1}{3}\right|\cdot\left(x^2+1\right)=0\)
mà \(x^2+1>=1\forall x\)
nên \(\left|x+\dfrac{1}{3}\right|=0\)
=>\(x+\dfrac{1}{3}=0\)
=>\(x=-\dfrac{1}{3}\)
c: \(\dfrac{16}{625}=\left(\dfrac{5}{2}\right)^{-4}\)