Giải giùm mik câu 3 phần b vs ạ
Mik đang cần gấp pleaseeeeeeeee
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15:
a: \(\text{Δ}=\left(m^2-m+2\right)^2-4m^2\)
=(m^2-m+2-2m)(m^2-m+2+2m)
=(m^2+m+2)(m^2-3m+2)
=(m-1)(m-2)(m^2+m+2)
Để phương trình co hai nghiệm phân biệt thì (m-1)(m-2)(m^2+m+2)>0
=>(m-1)(m-2)>0
=>m>2 hoặc m<1
b: x1+x2=m^2-m+2>0 với mọi m
x1*x2=m^2>0 vơi mọi m
=>Phương trình luôn có hai nghiệm dương phân biệt
1 are
2 am
3 is
4 are
5 are
6 are
7 is
8 is
9 is
10 are
IV
1 is writing
2 are losing
3 is having
4 is staying
5 am not lying
6 is always using
7 are having
8 Are you playing
9 are not touching
10 Is - listening
11 Is- winning
12 am not staying
13 is not working
14 is not reading
15 isn't raining
16 am not listening
17 Are they making
18 Are you doing
19 Is - sitting
20 is - doing
21 are-putting
22 are-wearing
23 is-studying
2, am
3, is
4,are
5,are
6,are
7,is
8,is
9,is
10,are
IV
1,2,7 OK
3,is having
4,has stayed
5,am not lying
6,always uses
8,Are-playing
9,not to touch
10,Is-listening
11,Are-winning
12,am not staying
13,isn't working
14,isn't reading
15,isn't raining
16,am not listening
17,Are-making
18,Are-doing
19,Is-sitting
20,is-doing
21,do-putting
22,do-wear
23,is-studying
Gọi tam giác đề bài cho là ΔABC vuông tại A. Đặt \(\hat{B}=\alpha\)
Xét ΔABC vuông tại A có
\(sin\alpha=\sin B=\frac{AC}{BC};cos\alpha=cosB=\frac{BA}{BC}\)
\(\tan\alpha=\tan B=\frac{AC}{AB};\cot\alpha=\cot B=\frac{AB}{AC}\)
\(\tan\alpha=\frac{AC}{AB}=\frac{AC}{BC}:\frac{AB}{BC}=\frac{\sin\alpha}{cos\alpha}\)
\(\cot\alpha=\frac{AB}{AC}=\frac{AB}{BC}:\frac{AC}{BC}=\frac{cos\alpha}{\sin\alpha}\)
\(\tan\alpha\cdot\cot\alpha=\frac{\sin\alpha}{cos\alpha}\cdot\frac{cos\alpha}{\sin\alpha}=1\)
\(\sin^2\alpha+cos^2\alpha\)
\(=\left(\frac{AC}{BC}\right)^2+\left(\frac{AB}{BC}\right)^2=\frac{AC^2+AB^2}{BC^2}=\frac{BC^2}{BC^2}=1\)
One of the biggest environmental impacts of human activities is air pollution







giải giúp mik vs ạ mik đang cần gấp!!!
Câu 6:
a: \(\overrightarrow{AC}=\left(3;-3\right)\)
\(\overrightarrow{DB}=\left(4-x_D;1-y_D\right)\)
Để ACBD là hình bình hành thì \(\left\{{}\begin{matrix}4-x_D=3\\1-y_D=-3\end{matrix}\right.\Leftrightarrow D\left(1;4\right)\)