18.9x178.178+0.49-17.8x189.189

18.9x178.178+0.49-17.8x189.189


18.9x178.178+0.49-17.8x189.189=189x17.8189-17.8x189.189+0.49=(189x17.8+189x0.0189)-17.8x189.189+0.49=17.8x(189-189.189)+17.8x0.0189+0.49=17.8x0.0189-17.8x0.0189+0.49=0.49



What is the graph represented by equation (X & # 178; - 4) &# 178; + (Y & # 178; - 4) &# 178; = 0?


The square of a number should be the square of the number, which is greater than or equal to 0. So only when x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(1); =The figure represented by 0 is the four vertices of a square



The standard deviation of data 5015025035045055506507508509 is______ .


The average of this set of data = (501 + 502 + 503 + 505 + 506 + 507 + 508 + 509) △ 9 = 2, variance = 19 [(501-505) 2 + (502-505) 2 + (503-505) 2 + (504-505) 2 + (505-505) 2 + (506-505) 2 + (507-505) 2 + (508-505) 2 + (509-505) 2] = 203, standard deviation = 203



How to calculate 501 + 502 + 503 + 504 + 505 + 506 + 507 by multiplication


504×7=3528