Let's find the formula for a rectangle:
2L+2W= 90
Now we know the length is equal to 6 more than 2 times the width, let's make the equation.
L=2w+6
Let's plug that in for L in the first equation.
2(2w+6)+2w=90
4w+12+2w
6w+12=90
6w=78
78÷6=13=w
The width is 13, but we need the length. We know the length is equal to 6 more than 2 times the width.
13×2= 26+6= 32
So, the length of the rectangle is 32 ft.
Answer:
B and D
Step-by-step explanation:
To determine which point is a solution, substitute the x- coordinate into the right side of the equation and if the value obtained agrees with the y- coordinate then it is a solution.
(5, 10)
y = 5² - 2(5) + 1 = 25 - 10 + 1 = 16 ≠ 10 ← not a solution
(- 1, 3)
y = (- 1)² - 2(- 1) + 1 = 1 + 2 + 1 = 3 ← This is a solution
(1, 2)
y = 1² - 2(1) + 1 = 1 - 2 + 1 = 0 ≠ 2 ← not a solution
(3, 4)
y = 3² - 2(3) + 1 = 9 - 6 + 1 = 4 ← this is a solution
Answer:
(2,1)
Step-by-step explanation:
We can see that the given triangle.
The coordinates of A are (-1,5) .
The coordinates of B are (-1,-3).
The coordinates of C are (5,-3).
Distance formula: ![\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
AB=
units
BC=
units
AC=
units
Pythagoras theorem:![(Hypotenuse)^2=(Base)^2+(Perpendicular\;side)^2}](https://tex.z-dn.net/?f=%28Hypotenuse%29%5E2%3D%28Base%29%5E2%2B%28Perpendicular%5C%3Bside%29%5E2%7D)
![AB^2+BC^2=8^2+6^2=100](https://tex.z-dn.net/?f=AB%5E2%2BBC%5E2%3D8%5E2%2B6%5E2%3D100)
![AC^2=(10)^2=100](https://tex.z-dn.net/?f=AC%5E2%3D%2810%29%5E2%3D100)
Therefore, ![AC^2=AB^2+BC^2](https://tex.z-dn.net/?f=AC%5E2%3DAB%5E2%2BBC%5E2)
When a triangle satisfied the Pythagoras theorem then, the triangle is right triangle.
Hence, the given triangle is a right triangle.
We know that circum-center of right triangle is the mid point of hypotenuse.
Mid-point formula:![x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%2Cy%3D%5Cfrac%7By_1%2By_2%7D%7B2%7D)
Using this formula then, we get
Mid-point of hypotenuse AC is given by
![x=\frac{-1+5}{2}=2,y=\frac{5-3}{2}=1](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-1%2B5%7D%7B2%7D%3D2%2Cy%3D%5Cfrac%7B5-3%7D%7B2%7D%3D1)
Hence, the circum-center of triangle is (2,1).