a) Ta có: \(\widehat{xOt}+\widehat{tOy}=90^0\)(hai góc phụ nhau)

\(\Leftrightarrow\widehat{tOy}=90^0-\widehat{xOt}=90^0-40^0\)

hay \(\widehat{tOy}=50^0\)

Vậy: \(\widehat{tOy}=50^0\)

Bài 5:

a) \(0,24\cdot-\dfrac{15}{4}=\dfrac{6}{25}\cdot-\dfrac{15}{4}=\dfrac{6\cdot-15}{25\cdot4}=-\dfrac{90}{100}=-\dfrac{9}{10}\)

b) \(4,5\cdot\dfrac{-4}{9}=\dfrac{9}{2}\cdot\dfrac{-4}{9}=\dfrac{9\cdot-4}{2\cdot9}=-\dfrac{4}{2}=-2\)

c) \(3,5\cdot-1\dfrac{2}{5}=\dfrac{7}{2}\cdot\dfrac{-7}{5}=\dfrac{7\cdot-7}{2\cdot5}=-\dfrac{49}{10}\)

Bài 6: 

a) \(1\dfrac{1}{17}\cdot\left(-2\dfrac{1}{8}\right)=\dfrac{18}{17}\cdot\dfrac{-17}{8}=\dfrac{18\cdot-17}{17\cdot8}=\dfrac{-18}{8}=-\dfrac{9}{4}\)

b) \(\left(-2\dfrac{1}{3}\right)\cdot1\dfrac{1}{14}=-\dfrac{7}{3}\cdot\dfrac{15}{14}=\dfrac{-7\cdot15}{3\cdot14}=-\dfrac{5}{2}\)

c) \(1,25\cdot\left(-3\dfrac{3}{8}\right)=\dfrac{5}{4}\cdot-\dfrac{27}{8}=\dfrac{5\cdot-27}{4\cdot8}=-\dfrac{135}{32}\)

`@` `\text {Ans}`

`\downarrow`

`5,`

`a)`

\(0,24\cdot\dfrac{-15}{4}\)

`= (0,24*(-15))/4`

`= (-3,6)/4 = -9/10`

`b)`

\(4,5\cdot\left(-\dfrac{4}{9}\right)\)

`= (4,5*(-4))/9`

`= -18/9 = -2`

`c)`

\(3,5\cdot\left(-1\dfrac{2}{5}\right)\)

`= 3,5*(-7/5)`

`= (3,5*(-7))/5`

`=(-24,5)/5 = -49/10`

`6,`

`a)`

\(1\dfrac{1}{17}\cdot\left(-2\dfrac{1}{8}\right)\)

`= 18/17 * (-17/8)`

`= -18/8 = -9/4`

`b)`

\(\left(-2\dfrac{1}{3}\right)\cdot1\dfrac{1}{14}\)

`= -7/3*15/14`

`= -1*5/2`

`= -5/2`

`c)`

\(1,25\cdot\left(-3\dfrac{3}{8}\right)\)

`= 1,25*(-27/8)`

`= (1,25*(-27))/8`

`= (-33,75)/8 = -135/32`

a. f(\(\dfrac{-1}{2}\)) = \(4.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)-2\) 

               = \(4.\dfrac{1}{4}-\left(\dfrac{-3}{2}\right)-\dfrac{4}{2}\)

               = \(\dfrac{2}{2}+\dfrac{3}{2}-\dfrac{4}{2}\)

               = \(\dfrac{1}{2}\)

Bài 3:

a) \(\left(-\dfrac{2}{3}\right)^2\cdot\left(\dfrac{2}{3}\right)^5\)

\(=\left(\dfrac{2}{3}\right)^2\cdot\left(\dfrac{2}{3}\right)^5\)

\(=\left(\dfrac{2}{3}\right)^{2+5}\)

\(=\left(\dfrac{2}{3}\right)^7\)

b) \(\left(-\dfrac{1}{2}\right)^5\cdot\left(-\dfrac{1}{2}\right)^3\)

\(=\left(-\dfrac{1}{2}\right)^{5+3}\)

\(=\left(-\dfrac{1}{2}\right)^8\)

\(=\left(\dfrac{1}{2}\right)^8\)

c) \(\left(\dfrac{6}{5}\right)^7\cdot\left(-\dfrac{6}{5}\right)^4\)

\(=\left(\dfrac{6}{5}\right)^7\cdot\left(\dfrac{6}{5}\right)^4\)

\(=\left(\dfrac{6}{5}\right)^{7+4}\)

\(=\left(\dfrac{6}{5}\right)^{11}\)

Bài 4:

a) \(\left(\dfrac{3}{7}\right)^4:\left(-\dfrac{3}{7}\right)^2\)

\(=\left(\dfrac{3}{7}\right)^4\cdot\left(\dfrac{3}{7}\right)^2\)

\(=\left(\dfrac{3}{7}\right)^{4+2}\)

\(=\left(\dfrac{3}{7}\right)^6\)

b) \(\left(\dfrac{5}{9}\right)^{11}:\left(\dfrac{5}{9}\right)^7\)

\(=\left(\dfrac{5}{9}\right)^{11-7}\)

\(=\left(\dfrac{5}{9}\right)^4\)

c) \(\left(\dfrac{2}{13}\right)^7:\left(\dfrac{2}{13}\right)^5\)

\(=\left(\dfrac{2}{13}\right)^{7-5}\)

\(=\left(\dfrac{2}{13}\right)^2\)

Bài 5:

a) \(\left(\dfrac{2}{3}\right)^0\cdot\left(\dfrac{2}{3}\right)^5\)

\(=1\cdot\left(\dfrac{2}{3}\right)^5\)

\(=\left(\dfrac{2}{3}\right)^5\)

b) \(\left(\dfrac{3}{5}\right)^7\cdot\left(\dfrac{3}{5}\right)^8\)

\(=\left(\dfrac{3}{5}\right)^{7+8}\)

\(=\left(\dfrac{3}{5}\right)^{15}\)

c) \(\left(-\dfrac{2}{7}\right)^9\cdot\left(-\dfrac{2}{7}\right)^{11}\)

\(=\left(-\dfrac{2}{7}\right)^{9+11}\)

\(=\left(-\dfrac{2}{7}\right)^{20}\)

\(400906=4\times100000+9\times100+6\)

\(abcde=a\times10000+b\times1000+c\times100+d\times10+e\)

\(#Wendy.Dang\)

đề yêu cầu j vậy bn

Viết PTHH á cậu?

PTHH:Zn+H₂O->ZnO+H₂

PTHH:Fe+H₂O->FeO+H₂

Số đo của \(\widehat{a}\) là:

180^o :3 .1 = 60^o

Vì \(\widehat{a}\) và  \(\widehat{b}\) kề bù nên tổng số đo hai góc là 180^o.

Số đo của \(\widehat{b}\) là:

180^o-60^o= 120^o

Vậy \(\widehat{b}\) = 120^o

Số đo của ˆaa^ là:

180^o :3 .1 = 60^o

Vì ˆaa^ và  ˆbb^ kề bù nên tổng số đo hai góc là 180^o.

Số đo của ˆbb^ là:

180^o-60^o= 120^o

Vậy ˆbb^ = 120^o

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