III gimnazjum

Matematyka 2001

100x - cena komputera x - cena kalkulatora   3⋅100x+20⋅x=12 800   300x+20x=12 800   320x=12 800   ∣:320   x=32012 800​=   321280​=   16640​=40   100x=100⋅40=4000  <section class="answer"><h4 class="answer-heading">Odpowiedź:</h4> Kalkulator kosztował 40 zł, a komputer ksztował 4000 zł. </section>

Rozwiązanie 1

Obliczamy powierzchnię pól uprawnych:  80%⋅2000 ha=0,8⋅2000 ha=1600 ha

19x - węgiel 3x - wodór 3x - tlen   19x+3x+3x=12,25   25x=12,25   ∣:25

v -prędkość pieszego   Pieszy i rowerzysta pokonali taką samą drogę - można ja obliczyć jako iloczyn prędkości i czasu.    Pieszy wyruszył o 15:00 i spotkał rowerzystę o 20:00, więc szedł przez 5 godzin.  Rowerzysta wyjechał o 17:30 i spotkał pieszego o 20:00, więc jechał przez 2,5 godziny.    5⋅v=2,5⋅16   ∣:5

x - pierwsza liczba y - druga liczba   {x+y=600   ∣−x130%⋅x+70%⋅y=480​   {y=600−x1.3x+0.7y=480​

x - cyfra dziesiątek liczby y - cyfra jedności liczby 10x+y - ta liczba x+y - suma cyfr tej liczby   {10x+y=7(x+y)x=y+3​ {10x+y=7x+7y  ∣−7x−yx=y+3​

Przypomnijmy sobie długości boków trójkąta prostokątnego o kątach 90°, 60° i 30°:   <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/29061/dZYZwwYwIa2X25KhCB37iQ%3D%3D_160010711d74d2c0360a4e2e92b1898b488895379d2ffd4b440148cdd3176990.jpg" width="276" height="183">   b3<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​=7 cm

a)   <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/30667/rI5N%2BeIFFy235bhaSJQQng%3D%3D_dd74c28f62082b79335ca899fcc37254b28dc2c497b55aac015b04fd3249dbff.jpg" width="624" height="317"> Dla przypomnienia - długości boków w trójkącie prostokątnym o kątach 90°, 60°, 30°: <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/29061/dZYZwwYwIa2X25KhCB37iQ%3D%3D_160010711d74d2c0360a4e2e92b1898b488895379d2ffd4b440148cdd3176990.jpg" width="247" height="164"> b=6 cm   b3<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​=63<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​ cm   - połowa drugiej przekątnej rombu   Druga przekątna tego rombu ma więc długość:  2⋅63<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​ cm=123<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​ cm

Obliczamy długość dłuższej podstawy:  3⋅6 cm=18 cm     <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/30669/ZFYWbIKmlVSV7MhFdNwPUA%3D%3D_2335ec59cd85846ea1a11505e4ec4febf40e0ff031f64fe1a4d7fd8324a96102.jpg" width="355" height="248"> Przypomnijmy sobie długości boków w trójkącie prostokątnym o katach 90°, 45°, 45°:   <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/22340/f6gfU1KyPAE0gXAZKqnvuw%3D%3D_b83fbd8cdf0f4eafda0aefebce679a097ea552834fc1724fc7984f6d2ec61058.jpg"> U nas:  b=12 cm   h=b=12 cm