Kopertina e librit Matematika 10 - 11: Pjesa II

Zgjidhja e ushtrimit 5

Zgjidhja e ushtrimit 5 të mësimit 6.5A në librin Matematika 10 - 11: Pjesa II nga shtëpia botuese Botime Pegi me autorë Steve Fearnley, June Haighton, Steve Lomax, Peter Mullarkey, James Nicholson dhe Matt Nixon.

Pyetja

Ekuacioni i rrethit është x2+y2=25x^2+y^2=25. Gjeni koordinatat e pikave, në qoftë se:

  1. x=0x=0
  2. y=4y=4
  3. y=−3y=-3
  4. x=5x=5
  5. y=2y=2
  6. x=−1x=-1
  7. x=12x=\dfrac{1}{2}
  8. y=−23y=-\dfrac{2}{3}

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Zgjidhja

  1. 02+y2=25→y2=25→y=25=50^2+y^2=25\rarr y^2=25\rarr y=\sqrt{25}=5
  2. x2+42=25→x2+16=25→x2=25−16→x2=9→x=9=3x^2+4^2=25\rarr x^2+16=25\rarr x^2=25-16\rarr x^2=9\rarr x=\sqrt{9}=3
  3. x2+(−3)2=25→x2+9=25→x2=25−9→x2=16→x=16=4x^2+(-3)^2=25\rarr x^2+9=25\rarr x^2=25-9\rarr x^2=16\rarr x=\sqrt{16}=4
  4. 52+y2=25→25+y2=25→y2=25−25→y2=0→y=05^2+y^2=25\rarr25+y^2=25\rarr y^2=25-25\rarr y^2=0\rarr y=0
  5. x2+22=25→x2+4=25→x2=25−4→x2=21→x=21x^2+2^2=25\rarr x^2+4=25\rarr x^2=25-4\rarr x^2=21\rarr x=\sqrt{21}
  6. (−1)2+y2=25→1+y2=25→y2=25−1→y2=24→y=24=4×6=4×6=26(-1)^2+y^2=25\rarr 1+y^2=25\rarr y^2=25-1\rarr y^2=24\rarr y=\sqrt{24}=\sqrt{4\times6}=\sqrt{4}\times\sqrt{6}=2\sqrt{6}
  7. (12)2+y2=25(\dfrac{1}{2})^2+y^2=25 →\rarr 14+y2=25\dfrac{1}{4}+y^2=25 →y2=25−14\rarr y^2=25-\dfrac{1}{4} →y2=100−14\rarr y^2=\dfrac{100-1}{4} →y2=994→\rarr y^2=\dfrac{99}{4}\rarr y=994y=\sqrt{\dfrac{99}{4}} =994=\dfrac{\sqrt{99}}{\sqrt{4}} =9×112= \dfrac{\sqrt{9\times11}}{2} =9×112= \dfrac{\sqrt{9}\times\sqrt{11}}{2} =3112= \dfrac{3\sqrt{11}}{2}
  8. x2+(−23)2=25x^2+(-\dfrac{2}{3})^2=25 →x2+49=25→\rarr x^2+\dfrac{4}{9}=25\rarr x2=25−49→x^2=25-\dfrac{4}{9}\rarr x2=225−49→x^2=\dfrac{225-4}{9}\rarr x2=2219→x^2=\dfrac{221}{9}\rarr x=2219x=\sqrt{\dfrac{221}{9}} =2219=\dfrac{\sqrt{221}}{\sqrt{9}} =2213=\dfrac{\sqrt{221}}{3}