a ( -o, 75+ 1/2) ;4/3
b 7/5 , 7/4 -32/5
tim x
10/3,x + 67/4= -13,25
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góc C=180-75-45=60 độ
Xét ΔABC có AB/sinC=AC/sinB
=>AB/sin60=2/sin45
=>\(AB=\sqrt{6}\)
Ta có:
\(\widehat{C}=180^o-75^o-45^o=60^o\)
Xét tam giác ABC ta có:
\(\dfrac{AB}{sinC}=\dfrac{AC}{sinB}\)
\(\Rightarrow AB=\dfrac{ACsinC}{sinB}\)
\(\Rightarrow AB=\dfrac{2\cdot sin60^o}{sin45^o}\)
\(\Rightarrow AB=\sqrt{6}\)
Vậy: ...
\(A=cos^212+sin^2\left(90-78\right)+cos^21+sin^2\left(90-89\right)\)
\(=cos^212+sin^212+cos^21+sin^21=1+1=2\)
\(B=sin^23+sin^287+sin^215+sin^275\)
\(=sin^23+cos^23+sin^215+cos^215=1+1=2\)
a. (75 + 45) : 5 = ?
Cách 1: (75 + 45) : 5
= 120 : 5 = 24
Cách 2: (75 + 45) : 5
= 75 : 5 + 45 : 5
= 15 + 9
= 24
b.(88 – 32) : 8 = ?
Cách 1: (88 - 32) : 8
= 56 : 8 =7
Cách 2: (88 - 32) : 8
= 88 : 8 – 32 : 8
= 11 – 4
= 7
\(a,=6\sqrt{3}-10\sqrt{3}+15\sqrt{3}=11\sqrt{3}\\ b,=2\sqrt{5}-\sqrt{5}+70\sqrt{5}=71\sqrt{5}\\ c,=\dfrac{\sin43^0}{\sin43^0}+1=1+1=2\\ d,Sửa:\dfrac{\tan32^0}{\cot68^0}-\cos30^0-\dfrac{\sin18^0}{\sin82^0}=\dfrac{\tan32^0}{\tan32^0}-\dfrac{\sqrt{3}}{2}-\dfrac{\sin18^0}{\cos18^0}=1-1-\dfrac{\sqrt{3}}{2}=-\dfrac{\sqrt{3}}{2}\)
\(n_{NH_4NO_2}=\dfrac{80}{64}=1.25\left(mol\right)\)
=> \(n_{NH_4NO_3\left(pư\right)}=\dfrac{1,25.75}{100}=0,9375\left(mol\right)\)
PTHH: NH4NO2 -to-> N2 + 2H2O
_____0,9375------>0,9375
=> VN2 = 0,9375.22,4 = 21(l)
a. -3/16
b. -79/20
tìm x
x=-9
1) a) \(\left(-0,75+\frac{1}{2}\right):\frac{4}{3}\)
\(=\left(-0,75+\frac{1}{2}\right)\cdot\frac{3}{4}\)
\(=\left(-0,75+0,5\right)\cdot0,75\)
\(=-0,25\cdot0,75\)
\(=-0,1875\)
Vậy: \(\left(-0,75+\frac{1}{2}\right):\frac{4}{3}=-0,1875\)
b) \(\frac{7}{5}\cdot\frac{7}{4}-\frac{32}{5}\)
\(=\frac{49}{20}-\frac{32}{5}\)
\(=\frac{49}{20}-\frac{128}{20}\)
\(=-\frac{79}{20}=-3\frac{19}{20}\)
Vậy: \(\frac{7}{5}\cdot\frac{7}{4}-\frac{32}{5}=-3\frac{19}{20}\)
2) \(\frac{10}{3}\cdot x+\frac{67}{4}=-13,25\)
<=> \(\frac{10}{3}\cdot x+\frac{67}{4}=-\frac{53}{4}\)
<=> \(\frac{10}{3}\cdot x=-\left(\frac{53}{4}+\frac{67}{4}\right)\)
<=> \(\frac{10}{3}\cdot x=-30\)
<=> \(x=-30:\frac{10}{3}\)
<=> \(x=-30\cdot\frac{3}{10}\)
<=> \(x=-9\)
Vậy: \(x=-9\)