Answer:
The coordinates are 3,1
Step-by-step explanation:
i did the assignment and it js correct
Answer:
1
Use the quadratic formula
=
−
±
2
−
4
√
2
x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}
x=2a−b±b2−4ac
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.
2
+
5
−
2
=
0
x^{2}+5x-2=0
x2+5x−2=0
=
1
a={\color{#c92786}{1}}
a=1
=
5
b={\color{#e8710a}{5}}
b=5
=
−
2
c={\color{#129eaf}{-2}}
c=−2
=
−
5
±
5
2
−
4
⋅
1
(
−
2
)
√
2
⋅
1
Step-by-step explanation:
this should help
Use Pythagorean Theorem:
a² + b² = c²
"a" is one side of the right triangle.
Its length is: radius of larger circle minus radius of smaller circle: (11 - 3 = 8)
The distance between the center of the circles creates the hypotenuse (17)
8² + b² = 17²
b² = 225
b = 15
Answer: the length of the common external tangent is 15.
