f(1)=3

f(-1)=3

\(\text{f(1)=}2.1^2+1=3\)

\(\text{f(-1)=}2.\left(-1\right)^2+1=3\)

\(\text{f(2)=}2.2^2+1=9\)

\(\text{f(0)=}2.0^2+1=1\)

\(\text{f(-3)=}=2.\left(-3\right)^2+1=19\)

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: \(\widehat{yOz}=40^0\)

a) Ta có: xOy+yOz=180⁰xOy+yOz=180⁰(hai góc kề bù)

              ⇔zOy+140⁰=180⁰

     hay yOz=40⁰yOz=40⁰

     Vậy: yOz=40⁰

1 were - would you play

2 weren't studying - would have 

3 had taken - wouldn't have got

4 would you go - could

5 will you give - is

6 recycle - won't be

7 had heard - wouldn't have gone

8 would you buy - had

9 don't hurry - will miss

10 had phoned - would have given

11 were - wouldn't eat

12 will go - rains

13 had known - would have sent

14 won't feel - swims

15 hadn't freezed - would have gone

giúp cái quái gì

giúp gì vậy bn

TL

Giúp gì vậy trời

Hok tốt

Ngày thứ hai nhà máy sản xuất được số tấn đường là:

1,17-0,12= 1,05 (tấn)

Cả hai ngày thứ nhất và thứ hai, nhà máy sản xuất được số tấn đương là:

1,17+1,05=2,22 (tấn)

Ngày thứ ba nhà máy sản xuất được số tấn đường là:

4,14-2,22=1,92 (tấn)

"Chúc em học tốt."

Bn cần bài nào trong 2 bài nhỉ?

e tách câu hỏi ra nhe tạm thời cj giúp mụt câu nhe

a: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=\dfrac{-5}{7}+1+\dfrac{5}{7}=1\)

b: \(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}\cdot4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{13}{56}\)

c: \(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{-24+45}{54}\cdot\dfrac{12}{7}\)

\(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{21}{54}\cdot\dfrac{12}{7}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{6+2}{9}=\dfrac{8}{9}\)

d: \(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{107}-\dfrac{1}{111}\)

\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{108}{333}=\dfrac{12}{37}\)

\(a,2\dfrac{1}{3}.5\dfrac{4}{7}=\dfrac{7}{3}.\dfrac{39}{7}=13\)

\(b,\dfrac{-1}{2}+\dfrac{4}{9}+\dfrac{3}{2}+\dfrac{5}{9}=-1+1=0\)

\(c,0,75+\left(-1,25\right)=-1,50\)

\(d,\dfrac{2}{5}\cdot\dfrac{3}{8}-\dfrac{2}{5}\cdot\dfrac{6}{11}=\dfrac{2}{5}\cdot\left(\dfrac{3}{8}-\dfrac{6}{11}\right)=\dfrac{2}{5}.\dfrac{-15}{88}=\dfrac{-3}{44}\)

1 C

2 B

3 B

4 C

5 B

6 B

7 D

8 B

9 A

10 B

Bài 4

1 for

2 of

3 about

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).