Answer:
The correct options are 1, 2 and 4.
Step-by-step explanation:
The given triangle is an equilateral triangle. All the sides of an equilateral triangle are same. All interior angles are equal with measure 60 degree.
The distance from the center of an equilateral triangle to the midpoint of a side is known as apothem. In the given figure letter a represents the apothem.
Point D is the midpoint of BC,
Use Pythagorean theorem in triangle COD. So, option 1 is correct.
The distance can not be negative, so the length of the apothem is approximately 2.5 cm. Option 4 is correct.
The line OC bisects the angle C.
Therefore option 2 is correct.
The perimeter of an equality triangle is
Option 3 is incorrect.
Option 5 is incorrect.
Therefore options 1, 2 and 4 are correct.
Answer:
Factor this polynomial:
F(x)=x^3-x^2-4x+4
Try to find the rational roots. If p/q is a root (p and q having no factors in common), then p must divide 4 and q must divide 1 (the coefficient of x^3).
The rational roots can thuis be +/1, +/2 and +/4. If you insert these values you find that the roots are at
x = 1, x = 2 and x = -2. This means that
x^3-x^2-4x+4 = A(x - 1)(x - 2)(x + 2)
A = 1, as you can see from equation the coefficient of x^3 on both sides.
Typo:
The rational roots can be
+/-1, +/-2 and +/-4
Step-by-step explanation:
Answer:
Ok
Step-by-step explanation:
Answer:
3
Step-by-step explanation:
I gussed but I'm not sure
Check the picture below, so, that'd be the square inscribed in the circle.
so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ 5}})\quad % (c,d) &({{ -8}}\quad ,&{{ -3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-8-(-2)]^2+[-3-5]^2} \\\\\\ d=\sqrt{(-8+2)^2+(-3-5)^2}\implies d=\sqrt{(-6)^2+(-8)^2} \\\\\\ d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%205%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%20-8%7D%7D%5Cquad%20%2C%26%7B%7B%20-3%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B-8-%28-2%29%5D%5E2%2B%5B-3-5%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-8%2B2%29%5E2%2B%28-3-5%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-6%29%5E2%2B%28-8%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B36%2B64%7D%5Cimplies%20d%3D%5Csqrt%7B100%7D%5Cimplies%20d%3D10)
that means the radius r = 5.
now, what's the center? well, the Midpoint of the diagonals, is really the center of the circle, let's check,
so, now we know the center coordinates and the radius, let's plug them in,
![\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{ k}})\qquad radius=&{{ r}}\\ &-5&1&5 \end{array} \\\\\\\ [x-(-5)]^2-[y-1]^2=5^2\implies (x+5)^2-(y-1)^2=25](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%7B%7B%20h%7D%7D%29%5E2%2B%28y-%7B%7B%20k%7D%7D%29%5E2%3D%7B%7B%20r%7D%7D%5E2%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Blllll%7D%0Acenter%5C%20%28%26%7B%7B%20h%7D%7D%2C%26%7B%7B%20k%7D%7D%29%5Cqquad%20%0Aradius%3D%26%7B%7B%20r%7D%7D%5C%5C%0A%26-5%261%265%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-5%29%5D%5E2-%5By-1%5D%5E2%3D5%5E2%5Cimplies%20%28x%2B5%29%5E2-%28y-1%29%5E2%3D25)
