Answer:
![23\sqrt{3}\ un^2](https://tex.z-dn.net/?f=23%5Csqrt%7B3%7D%5C%20un%5E2)
Step-by-step explanation:
Connect points I and K, K and M, M and I.
1. Find the area of triangles IJK, KLM and MNI:
![A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\](https://tex.z-dn.net/?f=A_%7B%5Ctriangle%20IJK%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20IJ%5Ccdot%20JK%5Ccdot%20%5Csin%20120%5E%7B%5Ccirc%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%202%5Ccdot%203%5Ccdot%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D%5Cdfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D%5C%20un%5E2%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20KLM%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20KL%5Ccdot%20LM%5Ccdot%20%5Csin%20120%5E%7B%5Ccirc%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%208%5Ccdot%202%5Ccdot%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D4%5Csqrt%7B3%7D%5C%20un%5E2%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20MNI%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20MN%5Ccdot%20NI%5Ccdot%20%5Csin%20120%5E%7B%5Ccirc%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%203%5Ccdot%208%5Ccdot%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D6%5Csqrt%7B3%7D%5C%20un%5E2%5C%5C%20%5C%5C%20%5C%5C)
2. Note that
![A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2](https://tex.z-dn.net/?f=A_%7B%5Ctriangle%20IJK%7D%3DA_%7B%5Ctriangle%20IAK%7D%3D%5Cdfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D%5C%20un%5E2%20%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20KLM%7D%3DA_%7B%5Ctriangle%20KAM%7D%3D4%5Csqrt%7B3%7D%5C%20un%5E2%20%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20MNI%7D%3DA_%7B%5Ctriangle%20MAI%7D%3D6%5Csqrt%7B3%7D%5C%20un%5E2)
3. The area of hexagon IJKLMN is the sum of the area of all triangles:
![A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2](https://tex.z-dn.net/?f=A_%7BIJKLMN%7D%3D2%5Ccdot%20%5Cleft%28%5Cdfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D%2B4%5Csqrt%7B3%7D%2B6%5Csqrt%7B3%7D%5Cright%29%3D23%5Csqrt%7B3%7D%5C%20un%5E2)
Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.
Answer:
40
Step-by-step explanation:
Answer:
256 diputados
Step-by-step explanation:
Paso 1
Encontramos la fracción de diputados que vota en blanco
Sea fracción total de diputados = 1
Fracción de diputados que vota en blanco = x
5/8 de los diputados votaron a favor.
= 1 - 5/8 = 3/8
3/4 del resto de diputados votaron en contra, de ahí:
= 3/4 de 3/8
= 9/32 votaron en contra
1 = 5/8 + 9/32 + x
x = 1 - (5/8 + 9/32)
x = 1 - (20+ 9/32)
x = 1 - 29/32
x = 3/32
Paso 2
Número total de diputados presentes =
3/32 × x = 24
3 veces / 32 = 24
Multiplicar en cruz.
3x = 32 × 24
x = 32 × 24/3
x = 256 diputados
De ahí que estuvieran presentes 256 diputados
Answer:
1. No because one can be 5 by 5 and the other can be 7 by 7
2. perimeter- 2·π·2.75≈17.27876 area-23.76
3. 1.38p=c
4. (m)(3600t)
Step-by-step explanation:
Idk how to explain fully
420 would be the middle number.
Hoped I Could Help:)