Check the picture below.
let's notice that the angle at K is an inscribed angle with an intercepted arc
![\bf \stackrel{\textit{using the inscribed angle theorem}}{K=\cfrac{\widehat{LI}+\widehat{IJ}}{2}}\implies 9x+1=\cfrac{(10x-1)+59}{2} \\\\\\ 9x+1=\cfrac{10x+58}{2}\implies 18x+2=10x+58\implies 8x+2=58 \\\\\\ 8x=56\implies x=\cfrac{56}{8}\implies x=7 \\\\[-0.35em] ~\dotfill\\\\ K=9x+1\implies K=9(7)+1\implies \boxed{K=64}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20the%20inscribed%20angle%20theorem%7D%7D%7BK%3D%5Ccfrac%7B%5Cwidehat%7BLI%7D%2B%5Cwidehat%7BIJ%7D%7D%7B2%7D%7D%5Cimplies%209x%2B1%3D%5Ccfrac%7B%2810x-1%29%2B59%7D%7B2%7D%20%5C%5C%5C%5C%5C%5C%209x%2B1%3D%5Ccfrac%7B10x%2B58%7D%7B2%7D%5Cimplies%2018x%2B2%3D10x%2B58%5Cimplies%208x%2B2%3D58%20%5C%5C%5C%5C%5C%5C%208x%3D56%5Cimplies%20x%3D%5Ccfrac%7B56%7D%7B8%7D%5Cimplies%20x%3D7%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20K%3D9x%2B1%5Cimplies%20K%3D9%287%29%2B1%5Cimplies%20%5Cboxed%7BK%3D64%7D)
now, let's notice something again, the angle at L is also an inscribed angle, intercepting and arc of 97 + 59 = 156, so then, by the inscribed angle theorem,
∡L is half that, or 78°.
now, let's take a look at the picture down below, to the inscribed quadrilateral conjecture, since ∡J and ∡I are both supplementary angles, then
∡I = 180 - 64 = 116°.
∡J = 180 - 78 = 102°.
You have to subtract the low from the high
20-(-25)
20+25
45
Let n be a number of balls: n = 14
In one pan you have 8 balls: 8n
In the other pan you have the rest of balls (14 - 8 = 6) along with a <span>weight of 20 grams = 6n + 20
So the pans are balanced:
8n = 6n + 20
Now, just solve the equation:
8n - 6n = 20
2n = 20
n = 20/2
n = 10
So, you know that a ball weights 10 grams.</span>
Answer:
El área de un rectángulo de largo L y ancho A es: L*A.
El área de un triangulo de alto H y base B es: H*B/2.
Ahora veamos las tres figuras:
Rectángulo amarillo:
Largo = 40m
Ancho = 45m
Área = 40m*45m = 1800 m^2
Rectángulo blanco:
Largo = 40m
Ancho = 35m
Área = 40m*35m = 1400m^2
Triangulo azul:
Base = 35m
Alto = 50m
Área = 35m*50m/2 = 875m^2
Area total = 1800 m^2 + 1400m^2 + 875m^2 = 4075m^2
