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Anonymous users2024-02-09Let t=root 1-x x=1-t 2 t>=0y=1-t 2+t=-(t-1 2) 2+5 4 When t=1 2 has a maximum value 5 4 The range is negative infinity, 5 4 The second formula: LG2+LG5=1
2lg5+2lg2+lg5)（2lg2+lg5+(lg2)^2)=(2+2lg5)（lg2+1+(lg2)^2)
LG5=1-LG2.
4-2lg2)（lg2+1+(lg2)^2)
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Anonymous users2024-02-08The square root of 1=y<=2.
LG25+2 3LG8+LG5 multiplied by LG20+LG squared 22LG5+2LG2+LG5*(2LG2+LG5)+LG2*LG2(LG2+LG5+2)(LG2+LG5).
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Anonymous users2024-02-07Question 1: Trigonometry commutation method; Note that one is x and the other is 1-x Question 2: Use the formula LGA+LGB=LGAB, LGA-LGB=LG(A B), LG(A N)=NLGA.
Original formula = 2LG5+2LG2+LG5(1+LG2)+(LG2) 22+LG5+LG2(LG5+LG2).
2+lg5+lg2
Note: LG5+LG2 LG(5*2)=LG10=1
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Anonymous users2024-02-06I choose DFrom the title 2c = 2 trace hole only 5 is formed by pf1 perpendicular pf2 to pf1f2 to form a long-angle triangle, so |pf1|of the square plus Zipei|pf2|The square is equal to the square of 2c, and then the syndicate|pf1|*|pf2|=2
It can be solved|pf2|of squared = 16, but due to |pf2|is a positive number for distance, so |pf2|=4, the same can be obtained|pf1|=2, which is also known from the definition of hyperbola|pf1|-|pf2|The absolute value of = 2a, and then according to the square of c = the square of a plus the square of b of the trembling. So b is squared to 4, and because the focus is f1(- 5,0).
f2 (50) is on the x-axis, so the answer is d
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Anonymous users2024-02-05Solution: Select C: (1 + 5) 2 + 2. Based on the three views, the geometry is a semi-conic (split in two from the vertices). The surface area consists of three parts: a semicircle on the base, half of the sector on the side of the cone, and an isosceles triangle that is divided into two.
S table S bottom + S curve + S 1* 5+ *2*2 (1+ 5) 2+2
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Anonymous users2024-02-04a, b, c three points are collinear, then there are 2a+b=1, so (2 a+1 b)=(2 a+1 b)*(2a+b)=4+2b a+2a b+1=5+2(b a+a b)>=5+4=9.
That is, the minimum value of 2 a+1 b is 9(Note a>0, b>0).
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Anonymous users2024-02-03(1)a5-a2=3d=-5-1=-6
d=-2a1=a2-d=1-(-2)=3
an=a1+(n-1)d=3+(-2)(n-1)=-2n+5sn=(a1+an)n 2=(3-2n+5)n 2=-n +4n The general formula for the series is an=-2n+5, and the sum of the first n terms is sn=-n +4n(2) The problem is misprinted, which should be the cn power of bn=2.
cn=(5-an)/2=[5-(-2n+5)]/2=nbn=2^(cn)=2ⁿ
b1=2b(n+1) bn=2 2 =2, is a fixed value series, which is a proportional series with 2 as the first term and 2 as the common ratio.
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