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The first question uses the induction formula, y=2*(1 2sinx+(root number 3) 2cosx)=2*(cos 60sinx+sin60cosx)=2sin(60+x).
The second question, y=2(sinx) 2+2sinxcosx=1-cos2x+sin2x
Supplemental disclosure: cos2x=(cosx) 2-(sinx) 2=1-2(sinx) 2=2(cosx) 2-1,sin2x=2sinxcosx
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Question 1 proposes 2, the original formula = 2 (1 sinx of 2 + 3cosx) = 2 (cos 60sinx + sin60cosx) = 2sin (60 + x).
Propose the square of 1 under the root number + the square of the root number 3.
Question 2, multiply, 2sin 2x+2sinxcosx=1-cos2x+sin2x
The power-reduced wide-angle formula is used here.
I'm a freshman in high school, so I won't be asked again.
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1) If the investment amount is x, the return of investing in stable products such as bonds is Y1, and the return of investing in risky products such as stocks is Y2, then.
y1=;y2 = root number x).
2) Set the maximum return to y, invest in risk-based products such as ** and other hidden potatoes x million yuan, then invest in stable products such as bonds and other stable products 30-x million yuan, so there is.
y=y1+y2=root number x).
root number x) +6
Root number(x)-2]+
When the root number (x) = 2, that is, x = 4, y has a maximum value, so when the investment in risk products such as ** is 40,000 yuan, then when the investment in stable products such as bonds is 260,000, there is a maximum return, which is 10,000 yuan.
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af=(2/3)a
First, find the length of Pa, let the length of Pa be the unknown number X, use the Pythagorean theorem to express Pb, PC, Pb multiplied by BC is equal to PC multiplied by Be, Pa can be found to find P, Pa can be found from the square of C, and then the square of Bc is equal to C multiplied by C, find Ce, CE is one-third of C, so that PE should be two-thirds of PC, and AF should be two-thirds of AB.
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You can start with the flip side, if there is no x r so that f(x)4
ps: In fact, this problem combines numbers and shapes, and it is also very easy to use function image solutions (the critical point is the tangent of a straight line and a parabola, and there is only one common point), so you might as well give it a try.
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The question was not copied correctly, check it.
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How did I learn ...... in junior high school?
Your first question doesn't seem to be right: a=1 b=-2a=-2 f(x)=x 2-2x+2 The second condition can find that the axis of symmetry is x=1 (it should be). >>>More
=a2-4b
1.The equation has no real solution and δ< 0 >>>More
Solution: Because a(n+1)+2sn 3=1 (1) so a(n+2)+2s(n+1) 3=1 (2) from (1)-(2) a(n+1)-a(n+2)=2a(n+1) 3a(n+1)=3a(n+2). >>>More
<>I'm sorry, I'm too watery.
Those are drafts. Do you need me to rewrite it? And I'm not sure about the accuracy ha. Guidance, guidance. >>>More
Let pc=b, the triangle abc becomes a, then ap= (a2-b 2), and rotate the bpc 60° counterclockwise around the point b'a, apparently p'bp=60,△bpc≌△bp'a, so bp'=bp >>>More