F=\dfrac{9}{5}C+32\\\\A.\\\dfrac{9}{5}C+32=F\ \ \ |-32\\\\\dfrac{9}{5}C=F-32\ \ \ |\cdot\dfrac{5}{9}\\\\C=\dfrac{5}{9}(F-32)
B.\\F=212\to C=\dfrac{5}{9}(212-32)=\dfrac{5}{9}\cdot180=100\\\\212^oF=100^oC
C.\\F=80\to C=\dfrac{5}{9}(80-32)=\dfrac{5}{9}\cdot48\approx26.7\\\\80^oF\approx26.7^oC
I pretty sure the answer is 134217728000000000000/17
The answer would be OC, most triangles, like right triangles have equivalent sides. 12, 12, 12in
Answer:
he area that is un-shaded is (x-3)(x-6).
The entire area is (2x+2)(3x-4).
Area of the shaded region is:
(2x+2)(3x-4) - (x-3)(x-6) =
(6x2-2x-8) - (x2-9x+18) =
6x2-2x-8-x2+9x-18 =
5x2+7x-26 hopes this help btw (;
