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a) Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)(hai góc kề bù)
\(\Leftrightarrow\widehat{zOy}+140^0=180^0\)
hay \(\widehat{yOz}=40^0\)
Vậy: \(\widehat{yOz}=40^0\)
\(\text{f(1)=}2.1^2+1=3\)
\(\text{f(-1)=}2.\left(-1\right)^2+1=3\)
\(\text{f(2)=}2.2^2+1=9\)
\(\text{f(0)=}2.0^2+1=1\)
\(\text{f(-3)=}=2.\left(-3\right)^2+1=19\)
1.A
2.A
3.B
4.C
5.B
6.C
7.A
8.A
9.B
10.A
11.B
12.A
13.C
14.B
15.B
16.A
17.A
18.A
19.A
20.C
a) Xét \(\Delta ABC:\)
\(BC^2=10^2=100.\\ AB^2+AC^2=6^2+8^2=100.\\ \Rightarrow BC^2=AB^2+AC^2.\)
\(\Rightarrow\Delta ABC\) vuông tại A (Pytago đảo).
\(\Rightarrow\widehat{BAC}=90^o.\)
b) Ta có: \(\widehat{B}+\widehat{D}=90^o.\\ \widehat{B}+\widehat{C}=90^o.\)
\(\Rightarrow\widehat{D}=\widehat{C}.\)
Xét \(\Delta ABC\) và \(\Delta AED:\)
\(\widehat{D}=\widehat{C}\left(cmt\right).\)
\(AC=AD\left(=8cm\right).\)
\(\widehat{BAC}=\widehat{EAD}\left(=90^o\right).\)
\(\Rightarrow\) \(\Delta ABC\) \(=\Delta AED\left(g-c-g\right).\)
c) Xét \(\Delta BDC:\)
DK là đường cao \(\left(DK\perp BC\right).\)
CA là đường cao \(\left(CA\perp AB\right).\)
Mà E là giao điểm của DK; CA (gt).
\(\Rightarrow\) E là trực tâm.\(\Rightarrow\) BE là đường cao.\(\Rightarrow\) \(BE\perp CD.\)b: \(x+1657-34524=1289\)
\(\Leftrightarrow x+1657=35813\)
hay \(x=34156\)
Kẻ đường cao AH cho tam giác ABC
sinB = AH/AB => \(\dfrac{\sqrt{3}}{2}=\dfrac{AH}{6}\Rightarrow AH=3\sqrt{3}\)cm
cosB = BH/AB => \(\dfrac{1}{2}=\dfrac{BH}{6}\Rightarrow BH=3cm\)
=> CH = BC - BH = 1 cm
Theo Pytago tam giác AHC vuông tại H
\(AC=\sqrt{AH^2+HC^2}=2\sqrt{7}cm\)
-> chọn A
17. The fire isn't very hot. It won't boil a kettle.
\(\Rightarrow\) The fire isn't hot enough to boil a kettle.
18. You are quite thin. You could slip between the bars.
\(\Rightarrow\) You are thin enough to slip between the bars.
19. He is very ill. He can't eat anything.
\(\Rightarrow\) He is too ill to eat anything.
20. Our new car is very wide. It won't get through those gates.
\(\Rightarrow\) Our new car is too wide to get through those gates.
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