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Anonymous users2024-02-10Solution: (1) If ab cd is known, then: bad=80°= adc is known to de-bisect adc
So: edc = 1 2 * adc
2) If AB CD is known, then: BCD=N°= ABC is known to be bisected ABC
So: ebc=1 2* abc
1/2*n°
Cross the point E, and make the parallel line EF of AB and CD to the left
Then: abe= bef= ebc=1 2*n°, edc= def=40°
So: bed= bef+ def
40+1/2*n°
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Anonymous users2024-02-091) A: ab cd
bad=∠adc
bad=80°
adc=∠bad=80°
and DE score ADC
edc=80÷2=40°
2) A: Let the intersection of DE and BC be F
edc=40° ∠bcd=n°
dfc=180-40-n=（140-n）°∠bfe=∠dfc=（140-n）°
ab//cd
abc=∠bcd=n°
and BE rated ABC
ebc=n÷2=（n/2）°
bed=180°-∠ebc-∠bfe=180°-（n/2）°-140-n）°=（40+n/2）°
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Anonymous users2024-02-08(lg5)^2+lg2*lg50
lg5)^2+lg2*lg2*25
lg5)^2+lg2*(lg2+lg25)=(lg5)^2+(lg2)^2+lg2*lg5^2=(lg5)^2+2lg2lg5+(lg2)^2=(lg5+lg2)^2
lg10)^2
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Anonymous users2024-02-07Solution 1: x 2-6x+8 x 2-6x+8+1-1 x 2-6x+9-1
x-3）^2-1^2
x-3-1)(x-3+1)
x-4）(x-2)
Solution 2: x 2+4x+3 x 2+4x+3+1-1=x 2+4x+4-1
x+2）^2-1^2
x+2-1)(x+2+1)
x+1)(x+3)
Since a>0 has an opening facing upwards and has a minimum value, we know that the two solutions of the function are x=-1 and x=-3, so we know that the axis of symmetry of the function is x=-2, so when the quadratic function has a minimum value ymin, the x value is -2
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Anonymous users2024-02-06x^2+4x+3=(x+2)^2-1
x+2)^2≥0
There is a minimum value in the above equation.
When x+2=0, i.e., x=-2, there is a minimum value of =-1
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Anonymous users2024-02-057. (1) The most pencils, the least two pens 3 + 2 = 5;
2) There are more apples and pears, 8-5 = 3 more.
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Anonymous users2024-02-04In two cases, 1 a b c a b c, the base liter buries b c
2 a b c a b c 180, resulting in a 90
So laugh at this choice c
f(x)=(ax+b) (x +1), in (- is the odd function.
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