Apply Pythagoras
h^(2) = x^(2) + y(2)
Hence for points ( -8,-3 ) & ( 4,5)
h^(2) = (-8-4)^(2) + ( -3-5)^(2).
h^(2) = (-12)^(2) + (-8)^(2)
h^(2) = 144 + 64
h^(2) = 208
h = sqrt(208)
h = 14.222051.... (The distance between the two points.
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The answer could be any number between 39 to 45. When doubled this produces 78 to 90. and with 7 subtracted this gives 71 to 83.
Possibility of two digit no whose sum is 11 29,38,47,56,65,74,83,92 Now subract 45 to each no mentioned above -16,-7,2,11,20,29,38,47 See after sixth comma 83 and 38. Reverse of 83 is 38. 38 is 45 less than 83. So 83 is original no
Possibility of two digit no whose sum is 11 Are 29,38,47,56,65,74,83,92 Subract 45 to each no mentioned above -16,-7,2,11,20,29,38,47 See after 6th comma 83 and 38 Reverse of 83 is 38. The no 38 is 45 less than 83. Original no is=83 and new no is 38
38
Between which two consecutive whole numbers does \sqrt{83} 83 lie? Fill out the sentence below to justify your answer and use your mouse to drag \sqrt{83} 83 to an approximately correct location on the number line.