8 szkoły podstawowej

Matematyka z kluczem 8

Wprowadźmy oznaczenie: x  -pierwsza z tych liczb y  -druga z tych liczb   Obliczmy, ile wynosi jedna z tych liczb:  80%x=10   0,8x=10   8x=100   x=12,5   Obliczmy wartość drugiej z tych liczb: 212,5+y​=10 ∣⋅2   12,5+y=20 ∣−12,5   y=7,5     Obliczmy jakim procentem liczby 7,5 jest liczba 10: 7,510​⋅100%=7510000​%=13331​%   Odp.: N ponieważ C.

Rozwiązanie 1

Sprawdźmy czy: 0,004⋅105<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.88em' viewBox='0 0 400000 1944' preserveAspectRatio='xMinYMin slice'><path d='M983 90
l0 -0
c4,-6.7,10,-10,18,-10 H400000v40
H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7
s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744
c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30
c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722
c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5
c53.7,-170.3,84.5,-266.8,92.5,-289.5z
M1001 80h400000v40h-400000z'/></svg>​&lt;20   0,004⋅100000<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.28em' viewBox='0 0 400000 1296' preserveAspectRatio='xMinYMin slice'><path d='M263,681c0.7,0,18,39.7,52,119
c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120
c340,-704.7,510.7,-1060.3,512,-1067
l0 -0
c4.7,-7.3,11,-11,19,-11
H40000v40H1012.3
s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232
c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1
s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26
c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z
M1001 80h400000v40h-400000z'/></svg>​=400<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>​=20   zatem:

Pierwsze zdanie jest prawdziwe ponieważ:

A. 6824x−5992=613  ∣+5992   6824x=6605  ∣:6824

Zauważmy, że: 1 km=1000 m   zatem:

Zauważmy, że z wierzchołka oznaczonego czerwoną kropką wychodzą 4 przekątne.