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A and B can find out and choose D.
Figure A 20 degrees + 60 degrees = 80 degrees, Figure B df ab, 180 degrees - 60 degrees - 40 degrees = 80 degrees, 120 degrees, Figure C 60 degrees, 60 degrees, late reading< adc = 180 degrees - 60 degrees = 120 degrees.
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Picture A: Yanyan Sui Chang AD handed over BC to E. ced= a+ b=20°+60°=80° (the outer angle is equal to the sum of the inner angles).
adc=∠c+∠ced=40°+80°=120°。(The outer angle is equal to the sum of the inner angles).
Figure B: Do DF AB at point D, and hand over BC to F.
cfd=∠b=60°;∠c=40°。
cdf=180°-60°-40°=80°。
adf=180°-∠a=180°-20°=160°。
adc=360°-∠adf-∠cdf=360°-160°-80°=120°。
(1+ BCD) + B + (2 + Round Socks BAD) = (1 + 2) + 40 ° + 60 ° + 20 ° = 1 + 2) + 120 ° = 180 °.
In ADC, ADC=180°-(1+2)=180°-60°=120°.
The ADC can be found for all three auxiliary lines, and the answer is D.
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A, B, and C can all be found to make full use of the sum of triangles 180 to solve.
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The perimeter is the circumference of a circle plus the length of the two sides of a rectangle, 64 2 100=.
The area is the area of the circle plus the area of the rectangle, (64 2) 2 +64 100=.
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This is a track and field field composed of two semicircles and straights, and the circumference of the half garden l== m).
Then the circumference of the entire track and field is.
l=2 r+2x100=m).
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If you're a human-taught version: 30
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11 questions choose D, 12 questions choose A, the analysis is as follows:
11.Let a=1, get y=x -2x-1
Let x=-1, get y=(-1) -2· (-1)-1=1+2-1=2≠1, a is false.
Let a=-2, get y=-2x +4x-1
Ream-2x +4x-1=0
2x -4x+1=0, discriminant formula =(-4) -4·2·1=8>0, the equation has two unequal real roots.
The function image has two different intersection points with the x-axis, and B is wrong.
y=ax -2ax-1=a(x-1) -a+1) quadratic coefficient a, axis of symmetry x=1
When a>0, the function image opening is upward, when x 1, y decreases with the increase of x, when c is wrong a<0, the function image opening is downward, when x 1, y increases with the increase of x, d is correct.
To sum up, you get: choose D
12.Let the right-angled side length of an isosceles right-angled triangle be a, and let the square side length be b, then s1 = a and s3 = b
The two right-angled sides of the other two right-angled triangles are a+b, a-bs2= (a+b)(a-b)= (a-b), parallelogram area=2s1+2s2+s3
2×½a²+2×½(a²-b²)+b²
a²+a²-b²+b²
2A 4S1 choose A
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Then, oe=of, oc=oa
Get the parallelogram AFCE
EF AC gets diamond-shaped AFCE
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