Answer:
11.31 units
Step-by-step explanation:
a^2 + b^2 = c^2
8^2 + 8^2 = c^2
64 + 64 = c^2
128 = c^2
Opposite of exponents is square roots, and we need to get rid of that exponent in c^2, meaning we find the square root.
= c
c = 11.31 units
The perimeter of an equilateral triangle is 6x-6.3. Draw and label a triangle to represent this situation. Then write an express
ion to represent its perimeter as a sum. Then write an expression to represent its perimeter as a product.
2 answers:
You might be interested in
So, we know the center is at -3,-1, ok
hmmm what's the radius anyway? well, we know that there's a point at 1,2 that is on the circle's path...hmmmm what's the distance from the center to that point? well, is the radius, let's check then.
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -3}}\quad ,&{{ -1}})\quad % (c,d) &({{ 1}}\quad ,&{{ 2}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ r=\sqrt{[1-(-3)]^2+[2-(-1)]^2}\implies r=\sqrt{(1+3)^2+(2+1)^2} \\\\\\ r=\sqrt{16+9}\implies r=\sqrt{25}\implies r=5](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%20-3%7D%7D%5Cquad%20%2C%26%7B%7B%20-1%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%201%7D%7D%5Cquad%20%2C%26%7B%7B%202%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%0Ar%3D%5Csqrt%7B%5B1-%28-3%29%5D%5E2%2B%5B2-%28-1%29%5D%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B%281%2B3%29%5E2%2B%282%2B1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B16%2B9%7D%5Cimplies%20r%3D%5Csqrt%7B25%7D%5Cimplies%20r%3D5)
so, what's the equation of a circle with center at -3, -1 and a radius of 5?
![\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{\quad k}})\qquad radius=&{{ r}}\\ &-3&-1&5 \end{array} \\\\\\\ [x-(-3)]^2+[y-(-1)]^2=5^2\implies (x+3)^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%5Cquad%20%20k%7D%7D%29%5Cqquad%20%0Aradius%3D%26%7B%7B%20r%7D%7D%5C%5C%0A%26-3%26-1%265%0A%5Cend%7Barray%7D%20%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-3%29%5D%5E2%2B%5By-%28-1%29%5D%5E2%3D5%5E2%5Cimplies%20%28x%2B3%29%5E2%2B%28y%2B1%29%5E2%3D25)
hmmm what's the radius anyway? well, we know that there's a point at 1,2 that is on the circle's path...hmmmm what's the distance from the center to that point? well, is the radius, let's check then.
so, what's the equation of a circle with center at -3, -1 and a radius of 5?