1задача
1) 5×4=20(м²)-отец и Саша
2задача
1)20:2=10(н)-ног
2)52:2=26(г)-голов
Найдите центр призмы со следующими вершинами: A(1,1,1), E(3,3,3), B(5,1,1), F(7,3,3), C(5,1,7), G(7,3,9) D(1,1,7), H(3,3,9).
Лариса83
Надо найти среднее арифметическое координат вершин
Х=(1+3+5+7+5+7+1+3)/8=4
Y=(1+3+1+3+1+3+1+3)/8=2
Z=(1+3+1+3+7+9+7+9)/8=5
ответ (4;2;5)
3 <span>3/5</span>·<span>1/3</span>=<span>3 · 5 + 3/5</span>·<span>1/3</span>=<span>18/5</span>·<span>1/3</span>=<span>18 · 1/5 · 3</span>=<span>18/15</span>=<span>15 · 1 + 3/15</span>=1 <span>3/15</span>=1 <span>1/<span>5
</span></span>1 <span>1/5</span>+3 <span>11/27</span>=<span>1 · 5 + 1/5</span>+<span>3 · 27 + 11/27</span>=<span>65</span>+<span>92/27</span>=27 · 6/27 · 5+<span>5 · 92/5 · 27</span>=162/135+<span>460/135</span>=162 + 460/135=<span>622/135</span>=135 · 4 + 82/135=4 <span>82/<span>135
</span></span>4 <span>82/135</span>-2 <span>17/18</span>=<span>4 · 135 + 82/135</span>-<span>2 · 18 + 17/18</span>=<span>622/135</span>-<span>53/18</span>=<span>2 · 622/2 · 135</span>-<span>15 · 53/15 · 18</span>=<span>1244/270</span>-<span>795270</span>=<span>1244 - 795/270</span>=<span>449/270</span>=<span>270 · 1 + 179/270</span>=1 <span>179/<span>270
</span></span><span>1 <span>179/270</span>:1 <span>23/27</span>=<span>1 · 270 + 179/270</span>:<span>1 · 27 + 23/27</span>=<span>449/270</span>:<span>50/27</span>=<span>449/270</span>·<span>27/50</span>=<span>449 · 27/270 · 50</span>=<span>12123/13500</span>=<span>449/500</span></span>
Если я не ошибаюсь,это модули.Значит |-10|-3 < |-10-3|