We have that
x²<span>-6x+7=0
</span>Group terms that contain the same variable
(x²-6x)+7=0
Complete the square Remember to balance the equation
(x²-6x+9-9)+7=0
Rewrite as perfect squares
(x-3)²+7-9=0
(x-3)²-2=0
(x-3)²=2
(x-3)=(+/-)√2
x=(+/-)√2+3
the solutions are
x=√2+3
x=-√2+3
Answer:
It is 1/3x wide
Step-by-step explanation:
add the terms, the answer is 1/3x
Answer: 80
Step-by-step explanation:
We can use the formula
, where
are the lengths of the diagonals.
By the distance formula,
![AC=\sqrt{(-2-2)^{2}+(-1-7)^{2}}=\sqrt{16+64}=4\sqrt{5}\\BD=\sqrt{(-8-8)^{2}+(-1-7)^{2}}=\sqrt{256+64}=8\sqrt{5}](https://tex.z-dn.net/?f=AC%3D%5Csqrt%7B%28-2-2%29%5E%7B2%7D%2B%28-1-7%29%5E%7B2%7D%7D%3D%5Csqrt%7B16%2B64%7D%3D4%5Csqrt%7B5%7D%5C%5CBD%3D%5Csqrt%7B%28-8-8%29%5E%7B2%7D%2B%28-1-7%29%5E%7B2%7D%7D%3D%5Csqrt%7B256%2B64%7D%3D8%5Csqrt%7B5%7D)
So, the area is ![\frac{(4\sqrt{5})(8\sqrt{5})}{2}=\boxed{80}](https://tex.z-dn.net/?f=%5Cfrac%7B%284%5Csqrt%7B5%7D%29%288%5Csqrt%7B5%7D%29%7D%7B2%7D%3D%5Cboxed%7B80%7D)
Answer:
368 sq. units.
Step-by-step explanation:
We have a square of side lengths 20 units and we cut four congruent isosceles right triangles from the corners of the square.
Now, the four isosceles right triangles have one leg equal to 4 units.
Therefore, the area of four triangles =
sq. units.
Now, we have the area of the given square is (20 × 20) = 400 sq. units.
Therefore, the area of the remaining octagon will be (400 - 32) = 368 sq. units. (Answer)
Answer:
7/9 = 42/n.
7n = 378,
n = 54.
Step-by-step explanation: