A (5, 4) and B( 5, -2) is 6 units
E(-2, -1) and F(-2, -5) is 4 units
G(3, -5) and H(6, -5) is 3 units
C(-4, 1) and D(1, 1) is 5 units
Answer:
Twelve-fifths.
Step-by-step explanation:
The diagram is shown below
Given triangle be right angled at
,
Withe reference to
,
Its opposite side is
, adjacent side is
and hypotenuse is ![AC=13](https://tex.z-dn.net/?f=AC%3D13)
We are to find the value of ![tan(A)](https://tex.z-dn.net/?f=tan%28A%29)
We know ![tan=\frac{opposite}{adjacent}](https://tex.z-dn.net/?f=tan%3D%5Cfrac%7Bopposite%7D%7Badjacent%7D)
Substituting the above values we get,
![tan(A)= \frac{BC}{AB}=\frac{12}{5}](https://tex.z-dn.net/?f=tan%28A%29%3D%20%5Cfrac%7BBC%7D%7BAB%7D%3D%5Cfrac%7B12%7D%7B5%7D)
So the answer is Twelve-fifths.
Answer:
<h2>0 - no solution</h2>
Step-by-step explanation:
![\text{Let}\\\\\left\{\begin{array}{ccc}ax+by=c\\dx+ey=f\end{array}\right\\\\\text{in its simplest form, i.e. that a and b, d and e are relatively first}](https://tex.z-dn.net/?f=%5Ctext%7BLet%7D%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dax%2Bby%3Dc%5C%5Cdx%2Bey%3Df%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Ctext%7Bin%20its%20simplest%20form%2C%20i.e.%20that%20a%20and%20b%2C%20d%20and%20e%20are%20relatively%20first%7D)
![\text{if}\ a=d,\ b=e,\ c=f,\ \text{then a system of equations has infinitely many solutions}\\\\\text{Example:}\\\\\left\{\begin{array}{ccc}x+3y=6\\-2x-6y=-12&\text{divide both sides by (-2)}\end{array}\right\\\\\left\{\begin{array}{ccc}x+3y=6\\x+3y=6\end{array}\right](https://tex.z-dn.net/?f=%5Ctext%7Bif%7D%5C%20a%3Dd%2C%5C%20b%3De%2C%5C%20c%3Df%2C%5C%20%5Ctext%7Bthen%20a%20system%20of%20equations%20has%20infinitely%20many%20solutions%7D%5C%5C%5C%5C%5Ctext%7BExample%3A%7D%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B3y%3D6%5C%5C-2x-6y%3D-12%26%5Ctext%7Bdivide%20both%20sides%20by%20%28-2%29%7D%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B3y%3D6%5C%5Cx%2B3y%3D6%5Cend%7Barray%7D%5Cright)
![\text{if}\ a=d,\ b=e,\ c\neqf,\ \text{then the system of equations has no solution}\\\\\text{Example:}\\\\\left\{\begin{array}{ccc}3x+9y=12&\text{divide both sides by 3}\\x+3y=1\end{array}\right\\\\\left\{\begin{array}{ccc}x+3y=4\\x+3y=1\end{array}\right](https://tex.z-dn.net/?f=%5Ctext%7Bif%7D%5C%20a%3Dd%2C%5C%20b%3De%2C%5C%20c%5Cneqf%2C%5C%20%5Ctext%7Bthen%20the%20system%20of%20equations%20has%20no%20solution%7D%5C%5C%5C%5C%5Ctext%7BExample%3A%7D%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7D3x%2B9y%3D12%26%5Ctext%7Bdivide%20both%20sides%20by%203%7D%5C%5Cx%2B3y%3D1%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B3y%3D4%5C%5Cx%2B3y%3D1%5Cend%7Barray%7D%5Cright)
![\text{In other cases it has one solution.}](https://tex.z-dn.net/?f=%5Ctext%7BIn%20other%20cases%20it%20has%20one%20solution.%7D)
![\text{We have:}\\\\\left\{\begin{array}{ccc}x+2y=2\\2x+4y=-8&\text{divide both sides by 2}\end{array}\right\\\\\left\{\begin{array}{ccc}x+2y=2\\x+2y=-4\end{array}\right\\\\\text{Conclusion:}\\\\\text{The system of equations has no solution.}](https://tex.z-dn.net/?f=%5Ctext%7BWe%20have%3A%7D%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B2y%3D2%5C%5C2x%2B4y%3D-8%26%5Ctext%7Bdivide%20both%20sides%20by%202%7D%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B2y%3D2%5C%5Cx%2B2y%3D-4%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Ctext%7BConclusion%3A%7D%5C%5C%5C%5C%5Ctext%7BThe%20system%20of%20equations%20has%20no%20solution.%7D)
![\text{Now I will show it}](https://tex.z-dn.net/?f=%5Ctext%7BNow%20I%20will%20show%20it%7D)
![\left\{\begin{array}{ccc}x+2y=2\\x+2y=-8&\text{change the sings}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}x+2y=2\\-x-2y=8\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad0=10\qquad\bold{FALSE}](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B2y%3D2%5C%5Cx%2B2y%3D-8%26%5Ctext%7Bchange%20the%20sings%7D%5Cend%7Barray%7D%5Cright%5C%5C%5Cunderline%7B%2B%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B2y%3D2%5C%5C-x-2y%3D8%5Cend%7Barray%7D%5Cright%7D%5Cqquad%5Ctext%7Badd%20both%20sides%20of%20the%20equations%7D%5C%5C.%5Cqquad0%3D10%5Cqquad%5Cbold%7BFALSE%7D)
13, 69, 89, 25, 55, 20, 99, 75, 42, 18, 66, 88, 89, 79, 75, 65, 25, 99, 66, 78. Order from least to greatest
Lana71 [14]
13, 18, 20, 25, 25, 42, 55, 65, 66, 66, 69, 75, 75, 78, 79, 88, 89, 95, 99
Answer:
L=135 W=105
Step-by-step explanation:
Your rectangle has two length sides and two width sides. That means two sides are 30 feet longer than the other sides. If you subtract 60 from 480, you get 420. Divide 420 by 4 and you get 105. So now you know the width sides are 105 feet long, so 105+105=210. Now all you have to do is add 30 to 105 and get 135. That is the length of the two longer sides. So add 135 and 135 to get 270. All you have to do is add 210 and 270 and you end up with the 480 feet of fence you were given. I know this is probably not the fancy way of doing it, but it works.