gibsonj338

$sin(x)<0.1$

${sin}^{-1}(sin(x))<{sin}^{-1}(0.1)$

$x<0.01001667421162...$

I just found another link that I think will help you:

$(x+y)\times2$

$2x+2y$

There are several ways to find the measure of angles:

01.  Get a protractor and roughly measure the angle to the nearest degrree.  This is asumed the angle is protrayed accurately.

02.  Use trigonometry

To use trigonometry, the angles have to form a triangle.  If you know at least one (two in some cases) sides  and at least one (two in some cases) angles, you can use either of two formulas depending on what information you are given.

Those two formulas are the Law of Sines and the Law of Cosines.

The formula for the Law of Sines:

$\frac{Sin(A)}{a}=\frac{Sin(B)}{b}=\frac{Sin(C)}{c}$ 

$\frac{a}{Sin(A)}=\frac{b}{Sin(B)}=\frac{c}{Sin(C)}$

A = angle A

B = angle B

C = angle C

a = side a

b = side b

c = side c

The formula for Law of Cosines:

a²=b²+c²-2bc*Cos(A)

b²=a²+c²-2ac*Cos(B)

c²=a²+b²-2ab*Cos(C)

A = angle A

B = angle B

C = angle C

a = side a

b = side b

c = side c

To understand the Law of Sines and Law of Cosines formulas, click on the following links:

$x\times y=324$

$x+y=-77$

There are several ways to solve this problem.  I will solve this problem this way,

Solve for x in the first equation

$x\times y=324$

$x\times \frac{y}{y}=\frac{324}{y}$

$x\times 1=\frac{324}{y}$

$x=\frac{324}{y}$

Subsitute $\frac{324}{y}$ for x in the second equations and solve for y.

$\frac{334}{y}+y=-77$

$\frac{334}{y}\times y+y\times y=-77\times y$

$\frac{334\times y}{y}+y\times y=-77\times y$

$334+y\times y=-77\times y$

$334+y^2=-77\times y$

$334+y^2=-77y$

$334+y^2+77y=-77y+77y$

$334+y^2+77y=0$

$y^2+77y+334=0$

$y^2+77y+334-334=0-334$

$y^2+77y+0=0-334$

$y^2+77y+0=-334$

$y^2+77y+0+1482.25 =-334+ 1482.25$

$y^2+77y+1482.25 =-334+ 1482.25$

$y^2+77y+1482.25 = 1148.25$

$(y+ 38.5 )^2= 1148.25$

$(\sqrt{y+ 38.5})^2=\sqrt{1148.25}$

$y+ 38.5=±\sqrt{1148.25}$

$y+ 38.5≈±\ 33.8858377497148505$

$y+ 38.5≈ 33.8858377497148505$ and $y+ 38.5≈- 33.8858377497148505$

$y+ 38.5-38.5≈ 33.8858377497148505-38.5$and $y+ 38.5-38.5≈ -33.8858377497148505-38.5$

$y+ 0≈ 33.8858377497148505-38.5$ and $y+ 0≈ -33.8858377497148505-38.5$

$y≈ 33.8858377497148505-38.5$ and $y≈ -33.8858377497148505-38.5$

$y≈-4.6141622502851495$ and $y≈-72.3858377497148505$

Subsitute y in the second equation and solve for x.

$x+(-4.6141622502851495)≈-77$ and $x+(-72.3858377497148505)≈-77$

$x-4.6141622502851495≈-77$ and $x-72.3858377497148505≈-77$

$x-4.6141622502851495+4.6141622502851495≈-77+4.6141622502851495$ and $x-72.3858377497148505+72.3858377497148505≈-77+72.3858377497148505$

$x-0≈-77+4.6141622502851495$ and $x-0≈-77+72.3858377497148505$

$x≈-77+4.6141622502851495$ and $x≈-77+72.3858377497148505$

$x≈-72.3858377497148505$ and $x≈-4.6141622502851495$

$\frac{1\frac{4}{7}}{1\frac{1}{4}}$

$\frac{1+\frac{4}{7}}{1\frac{1}{4}}$

$\frac{1+\frac{4}{7}}{1+\frac{1}{4}}$

$\frac{\frac{7}{7}+\frac{4}{7}}{1+\frac{1}{4}}$

$\frac{\frac{11}{7}}{1+\frac{1}{4}}$

$\frac{\frac{11}{7}}{\frac{4}{4}+\frac{1}{4}}$

$\frac{\frac{11}{7}}{\frac{5}{4}}$

$\frac{11}{7}\times\frac{4}{5}$

$\frac{44}{35}$

$1\frac{9}{35}$

$1.2571428571428571...$

Two ways to solve this.

The long way:

$(2+\frac{1}{2})^2$

$(2+\frac{1}{2})(2+\frac{1}{2})$

$2\times2+2\times\frac{1}{2}+2\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}$

$4+2\times\frac{1}{2}+2\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}$

$4+\frac{2}{2}+2\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}$

$4+1+2\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}$

$4+1+\frac{2}{2}+\frac{1}{2}\times\frac{1}{2}$

$4+1+1+\frac{1}{2}\times\frac{1}{2}$

$4+1+1+\frac{1}{4}$

$5+1+\frac{1}{4}$

$6+\frac{1}{4}$

$\frac{24}{4}+\frac{1}{4}$

$\frac{25}{4}$

$6\frac{1}{4}$

$6.25$

The short way:

$(2+\frac{1}{2})^2$

$(\frac{4}{2}+\frac{1}{2})^2$

$(\frac{5}{2})^2$

$\frac{5^2}{2^2}$

$\frac{25}{2^2}$

$\frac{25}{4}$

$6\frac{1}{4}$

$6.25$

$7\frac{1}{5}-3\frac{3}{4}$

$\frac{36}{5}-\frac{15}{4}$

$\frac{144}{20}-\frac{75}{20}$

$\frac{69}{20}$

$3\frac{9}{20}$

$3.45$

I can see where you got your answer of $1\frac{7}{9}$.  $3\frac{1}{3}$ is not $3\times\frac{1}{3}$, $3\frac{1}{3}$ is $3+\frac{1}{3}$. 

$\frac{2}{3}\times3\frac{1}{3}+\frac{1}{3}(3+\frac{1}{3})$

$\frac{2}{3}\times(3+\frac{1}{3})+\frac{1}{3}(3+\frac{1}{3})$

$\frac{2}{3}(3+\frac{1}{3})+\frac{1}{3}(3+\frac{1}{3})$

$(\frac{6}{3}+\frac{2}{9})+\frac{1}{3}(3+\frac{1}{3})$

$(\frac{18}{9}+\frac{2}{9})+\frac{1}{3}(3+\frac{1}{3})$

$\frac{20}{9}+\frac{1}{3}(3+\frac{1}{3})$

$\frac{20}{9}+\frac{3}{3}+\frac{1}{9}$

$\frac{20}{9}+\frac{9}{9}+\frac{1}{9}$

$\frac{29}{9}+\frac{1}{9}$

$\frac{30}{9}$

$3\frac{3}{9}$

$3\frac{1}{3}$

$\frac{2}{\frac{1}{5}}$

$2\times\frac{5}{1}$

$2\times5$

$10$