Giúp minh câu 9,10,11,12 với
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9 I am interested in watching cartoons
10 for cheap and clean, and effective source of energy
11 sgould switch off electrical appliances when they are not in use
12 in rural areas is lower than in cities
9 I am interested in watching cartoons
10 for cheap and clean, and effective source of energy
11 you switch off electrical appliances when they are not in use
12 in rural areas is lower than in cities
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17:
\(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\)
\(\Rightarrow\left|\overrightarrow{AC}\right|=a\) vì tam giác ABC đều
18.
\(\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}\)
\(\Rightarrow\left|\overrightarrow{ BA}\right|=2a\) vì tam giác ABC đều
19 .
\(\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}\)
\(\Rightarrow\left|\overrightarrow{CB}\right|=3a\) vì tam giác ABC đều
20.
\(\overrightarrow{BA}+\overrightarrow{AC}=\overrightarrow{BC}\)
\(\Rightarrow\left|\overrightarrow{BC}\right|=4a\) vì tam giác ABC đều
Mình chữa nốt 14, 15, 16, 21
14, \(\overrightarrow{BD}+\overrightarrow{CA}=\overrightarrow{BA}+\overrightarrow{CD}\)
\(\Leftrightarrow\) \(\overrightarrow{BD}-\overrightarrow{BA}=\overrightarrow{CD}-\overrightarrow{CA}\)
\(\Leftrightarrow\) \(\overrightarrow{AD}=\overrightarrow{AD}\) (luôn đúng)
15, \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}\)
\(\Leftrightarrow\) \(\overrightarrow{CA}-\overrightarrow{CB}=\overrightarrow{DA}-\overrightarrow{DB}\)
\(\Leftrightarrow\) \(\overrightarrow{BA}=\overrightarrow{BA}\) (luôn đúng)
16, \(\overrightarrow{DC}+\overrightarrow{AB}=\overrightarrow{DB}+\overrightarrow{AC}\)
\(\Leftrightarrow\) \(\overrightarrow{DC}-\overrightarrow{DB}=\overrightarrow{AC}-\overrightarrow{AB}\)
\(\Leftrightarrow\) \(\overrightarrow{BC}=\overrightarrow{BC}\) (luôn đúng)
21,
Ta có: \(\left|\overrightarrow{F}\right|=\left|\overrightarrow{F1}+\overrightarrow{F2}\right|\)
\(\Leftrightarrow\) \(F^2=F1^2+F2^2+2\overrightarrow{F1}\cdot\overrightarrow{F2}\)
\(\Leftrightarrow\) \(F^2=F1^2+F2^2+2F1\cdot F2\cdot cos\left(\overrightarrow{F1},\overrightarrow{F2}\right)\)
\(\Leftrightarrow\) \(F=\sqrt{500^2+500^2+2\cdot500\cdot500\cdot cos60^o}\)
\(\Leftrightarrow\) \(F\approx866\left(N\right)\)
Chúc bn học tốt!
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16.
\(A=\dfrac{16^2.16^3}{4^8}=\dfrac{4^4.4^6}{4^8}=\dfrac{4^{10}}{4^8}=4^2=16\)
\(B=\dfrac{8^2.8^3}{2^{11}}=\dfrac{2^6.2^9}{2^{11}}=\dfrac{2^{15}}{2^{11}}=2^4=16\)
17.
\(6,673\)
18.
\(\Rightarrow x=\dfrac{-3}{5}.15=-9\)
19.
\(x=12:\dfrac{3}{4}=16\)
20.Áp dụng t/c dtsbn ta có;
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{x+y}{8+12}=\dfrac{40}{20}=2\)
\(\dfrac{x}{8}=2\Rightarrow x=16\\ \dfrac{y}{12}=2\Rightarrow y=24\)
21.
Áp dụng t/c dtsbn ta có;
\(\dfrac{x}{4}=\dfrac{y}{9}=\dfrac{x-y}{4-9}=\dfrac{15}{-5}=-3\)
\(\dfrac{x}{4}=-3\Rightarrow x=-12\\ \dfrac{y}{9}=-3\Rightarrow y=-27\)
22.
gọi số học sinh nam, nữ lần lượt là a,b
Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{6}=\dfrac{b}{7}\\a+b=39\end{matrix}\right.\)
Áp dụng t/c dtsbn ta có;
\(\dfrac{a}{6}=\dfrac{b}{7}=\dfrac{a+b}{6+7}=\dfrac{39}{13}=3\)
\(\dfrac{a}{6}=3\Rightarrow a=18\\ \dfrac{b}{7}=3\Rightarrow b=21\)
Vậy ...
23.\(\sqrt{16}=4\)
24.\(\sqrt{x}=5\Rightarrow x=25\)
25.B
26.A
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Câu 9: B
Câu 10: A
Câu 11: B
Câu 12: C