giúp với nha mn
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b: Xét ΔABH vuông tại H có
\(AB^2=AH^2+HB^2\)
hay AH=12(cm)
Xét ΔAHB vuông tại H có
\(\sin\widehat{B}=\cos\widehat{C}=\dfrac{AH}{AB}=\dfrac{12}{13}\)
\(\cos\widehat{B}=\sin\widehat{C}=\dfrac{5}{13}\)
\(\tan\widehat{B}=\cot\widehat{C}=\dfrac{12}{5}\)
\(\cot\widehat{B}=\tan\widehat{C}=\dfrac{5}{12}\)
16 seeing
17 was sent
18 hadn't flied
19 being taken
20 saw
21 had gone
22 were these photos taken
23 did we discovered
24 to tell
25 would have gone
26 be put
27 to concentrate
28 spoke
29 confused
30 had he arrived
31 C
32 A
33 B
34 C
35 C
36 B
37 C
38 C
39 B
40 C
41 achivement
42 disappearance
43 popularity
44 passionate
45 opposition
bữa sau bạn nhớ giải thích nữa nha chớ mình không biết tại sao ra đáp án đó đâu
\(Ba\left(NO_3\right)_2+H_2SO_4\rightarrow2HNO_3+BaSO_4\downarrow\)
\(CaCO_3+HNO_3\rightarrow Ca\left(NO_3\right)_2+CO_2\uparrow+H_2O\)
\(3AgNO_3+H_3PO_4\rightarrow AgPO_4\downarrow+HNO_3\)
\(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(=\frac{\left(2^3\right)^5.5^8+\left(-1\right)^5.2^5.\left(2.5\right)^9}{2^{16}.5^7+\left(2^2.5\right)^8}\)
\(=\frac{2^{15}.5^8+\left(-1\right).2^5.2^9.5^9}{2^{16}.5^7+\left(2^2\right)^8.5^8}\)
\(=\frac{2^{15}.5^8+\left(-1\right).2^{14}.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{14}.5^8.\left(2-1-5\right)}{2^{16}.5^7.\left(1+5\right)}\)
\(=\frac{5.\left(-4\right)}{2^2.6}=\frac{-20}{24}=\frac{-5}{6}\)
Mình nhầm xíu :
\(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}=\frac{\left(2^3\right)^5.5^8+\left(-1\right)^5.2^5.\left(2.5\right)^9}{2^{16}.5^7+\left(2^2.5\right)^8}\)
\(=\frac{2^{15}.5^8+\left(-1\right).2^5.2^9.5^9}{2^{16}.5^7+\left(2^2\right)^8.5^8}\)\(=\frac{2^{15}.5^8+\left(-1\right).2^{14}.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{14}.5^8.\left(2-1.5\right)}{2^{16}.5^7.\left(1+5\right)}\)\(=\frac{5.\left(-3\right)}{2^2.6}=\frac{-15}{24}=\frac{-5}{8}\)