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Anonymous users2024-02-081) Let -x+8=k x, x -8x+k=0, require x to have 2 solutions, that is, 64-4k 0
So there is k 16 (k ≠ 0).
2) (1) can be solved x=4 16-k, respectively, the abscissa of a and b two points may wish to be set xa xb
then xa=4+ 16-k, ya=4- 16-k; xb=4-√16-k，yb=4+√16-k。
Drawing it is clear that at k 0, both points are in the first quadrant, aob 90°0 k 16, a in the fourth quadrant, b in the second quadrant, aob 90°
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Anonymous users2024-02-071) Simultaneous 2 analytic formulas -x+8=k x -x 2+8x-k=0 may be f(x)=-x 2+8x-k=-(x-4) 2+16-k Because there are 2 intersections, 16-k>0 k<16 and k≠0
2) If 00, then the angle AOB is less than 90
If k<0 then y=k x is in.
2. The intersection of the image of y=-x+8 must be in the second and fourth quadrants, then x1,x2 and y1,y2 are x1x2<0 y1y2<0 In the same way, cos angle aob=oa*ob (|oa||ob|)=(x1x2+y1y2)/(|oa||ob|<0 so the angular AOB is greater than 90
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Anonymous users2024-02-06<>2) Angular AOB less than 90
And only when the intersection point a b moves to the intersection point of the primary function and the coordinate axis is that the angle is 90 degrees.
Therefore, the angle AOB is less than 90 degrees.
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Anonymous users2024-02-05Time. 2.AOB = 2 times 90° (A and B are symmetrical with respect to the origin).
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Anonymous users2024-02-04Let the side length of the square be acm, and the rectangle in the upper left corner is xcm long, then its width will be (4 x) cm, according to the title.
The rectangle in the upper right corner is long (a-x) cm, wide (4 x) cm, area 4 (a-x) x=12, simplified to (a-x) x=3
The rectangle in the lower right corner is long (A-X) cm, wide (A-4 X) cm, area (A-X) (A-4 X) = 20, and the area of the rectangle is obtained by cracking = x (A-4 X) = 20 3cm 2
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Anonymous users2024-02-03As you can see from the title picture, this is a square divided into two rectangles, the number of closed and the small square. This diagram is often used to interpret the formula (a+b) 2=a 2+2ab+b 2.
The area of this rectangle is:
Suspicion 20 + 4) -12 = 12 (cm 2).
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Anonymous users2024-02-02A regular hexagon has six corners, each of which is 120°.
The corners of the six corners are bisected, and finally the six corners bisector will intersect one point inside the hexagon, and the hexagon will be divided into six regular triangles with sides equal to the length of the original hexagon.
Then according to Zheng Jian's sine and Sun staring cosine, the height of the triangle is three times the length of the side of the root number.
Then the area of the regular hexagon is determined to be three, multiplied by three times the length of the side.
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Anonymous users2024-02-01Untie. If the transportation volume of small trucks is x, then the total number of first transportation (large and small trucks) is 3x+4+x; The total number of second transports (large and small trucks) is 6x-5+x. If the tonnage of the two transports is the same, there is 3x+4+x=6x-5+x, and the solution is x=3
So the sum of the two transports is 2*(6x-5+x)=2*(6*3-5+3)=32 (tons).
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Anonymous users2024-01-31Solution, design: The first batch of large trucks transported x tons, and small trucks transported y tons.
3y+4=x
x+5=6y (the goods transported twice are equal).
x=13，y=5
13+5）x2=26
A: There are 26 tons of yellow sand.
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Anonymous users2024-01-30The tonnage of the first batch of large trucks transporting yellow sand is y tons, the tonnage of small trucks transporting yellow sand is x tons, the tonnage of the second batch of large trucks is y+5 tons, and the two batches of yellow sand have a total of 2y+2 (y+5) tons.
3x+4=y
6x=y+5
x=3,y=13
2y+2(y+5)=62
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Anonymous users2024-01-29The first big trucks carried X, and the small trucks carried Y
x=3y+4
x+5=6y
x=13,y=5
A total of 26 tons.
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Anonymous users2024-01-28Error above, the tonnage of the second pickup truck delivery is not y, it should be x+5 6
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Anonymous users2024-01-271 The total square area is.
The rectangular area is.
The square is divided into 169 small squares with side length, and 168 of them can be assembled into rectangles of length and width.
In a rectangle, each row has 8 small squares and each column has 21 small squares.
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Anonymous users2024-01-26square meters, change to squares? No way.
But there's an interesting trick up the sleeve, something like that.
I'll do a rough sketch and make do with it.
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Anonymous users2024-01-25No, the length is not enough.
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Anonymous users2024-01-24Knowing by solution x<4 9.
Inequality (4a-3b) x>2b-a
4a-3b<0
So x<(2b-a) (4a-3b)=4 9 so 4*(4a-3b)=9*(2b-a), 16a-12b=18b-9a, 25a=30b, a=6 5b
And 4a-3b=9 5b<0, so b<0
So the inequality ax>b is reduced to.
6/5bx>b
Divide the two sides by b.
6/5x<1
x<5/6
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Anonymous users2024-01-23x<5/6
It is easy to know that 4a-3b<0, (2b-a) (4a-3b)=4 9 gets 5a=6b, and gets b a=5 6, that is, 3b=(5 2)a, so 4a-(5 2)a<0, get a<0
So x
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Anonymous users2024-01-22Equal x is discussed separately for odd and even numbers, both are equal.
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Anonymous users2024-01-21As you can see, x is an odd number.
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Anonymous users2024-01-20For the function y=-x +2ax+5 (-1 x 1), the opening is downward, and the axis of symmetry is a
According to the position of the axis of symmetry, it is discussed in three categories through the combination of numbers and shapes:
1°If a<-1, then the function decreases monotonically on [-1,1], then the minimum value is f(1)=4+2a, and the maximum value is f(-1)=4-2a, 4+2a, 1,4-2a, 8
Solution: -3 2 a<-1
2° Ruo-1 a 1
Then the maximum value is f(a)=a +5, and the minimum value is f(1)=4+2a or f(-1)=4-2a
a +5 8,4+2a 1,4-2a 1 solves: -1 a 1
3° If a>1, then the function increases monotonically on [-1,1], then the maximum value is f(1)=4+2a, and the minimum value is f(-1)=4-2a, 4+2a, 8,4-2a1
Solution: 1 To sum up: the value range of a is: [-3 2,3 2] [Mathematics, Physics and Chemistry for Middle School Students] team will answer for you! I wish you progress in your studies, and you can ask if you don't understand!
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Anonymous users2024-01-19j solution: Let t=sinx, then -1 t 1
f(x) becomes, y=-t +2at+5=-(t-a) +5+a) 1 y 8
Axis of symmetry t=a. The opening is downward, monotonically increasing on the left and decreasing on the right.
When -1 t 1 a, i.e. a 1, on the left side, -1-2a+5 1, -1+2a+5 8 1 a
When a-1 t 1, i.e. a-1, right. -1-2a+5 8, -1+2a+5 1 do not exist.
When -1 a 1. When a+1=1-a, i.e., a=0, a +5 8, -a+2a+5 1 exists.
When a+1>1-a, i.e., a>o, -1-2a+5 1,a +5 8 0, when a+1<1-a, i.e., a<0, -1+2a+5 1,a+5 8<0
To sum up,
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Anonymous users2024-01-18Knowing the function f(x)=-sin x+2asinx+5, if x r, there is 1 f(x) 8, find the range of values of a.
Solution: f(x)=-(sin x-2asinx)+5=-[(sinx-a) -a ]+5=-(sinx-a) +a +5
When a=0, f(x)=-sin x+5, then 4 f(x) 5;Therefore a=0 is desirable.
When a>0, a 2 is obtained from maxf(x)=-(1-a) +a +5=2a+4 8
From minf(x)=-(-1-a) +a +5=-2a+4 1, we get a 3 2;
Therefore, when 0 is < 0, a -2 is obtained from maxf(x)=-(-1-a) +a +5=-2a+4 8
From minf(x)=-(-1-a) +a +5=-2a+4 1, we get a 3 2;
Therefore, -3 2 a<0 is desirable.
= is the range of values of a.
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Anonymous users2024-01-17
1, -1<=a<=1, the maximum value appears at x=a, and the minimum value is the smaller one between f(1) and f(-1).
2. When a<-1, the maximum value is f(1) and the minimum value is f(-1).
3. When a>1, the maximum value is f(-1) and the minimum value is f(1).
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Anonymous users2024-01-16
f(x)=-sin²x+2asinx+5
Let t=sinx [-1,1], then.
f(x)=-t +2at+5=-(t-a) +5+a =g(t), the g(t) image is a parabola with an opening pointing downward, the domain is defined as t [-1,1], the axis of symmetry is t=a, and the maximum value is 5+a
1) On x r, if there is 1 f(x) 8, i.e. 1 g(t) 8, then if -1 a 1, then the maximum value is 5 + a 8, and the solution is -3 a 3, and the intersection with the assumption is -1 a 1 If a 1, then g(t) is monotonically increasing on [-1,1], the minimum value of the function is g(-1)=-1-2a+5 1, and the maximum value is g(1)=-1+2a+5 8
The solution gives a 3 2, and a 2, and intersects with the assumptions to give 1 a 3 2 If a -1, then g(t) is a monotonically decreasing function on [-1,1], the minimum value is g(1)=-1+2a+5 1, and the maximum value is g(-1)=-1-2a+5 8
The solution gives a -3 2, and a -2, and the intersection with the hypothesis gives -3 2 a -1 In summary, if the value range is [1,8], then the value range of a is [-3 2,3 2](2)f(x)=0 has a real solution, that is, -(t-a) +5+a =0 has a solution on [-1,1].
If t1=a- (5+a), t2=a+ (5+a) then there is -1 a- (5+a) 1, or -1 a+ (5+a) 1, and the above inequality can be solved to obtain -2 a 2, or a -2 takes the intersection to get a 2
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Anonymous users2024-01-15
The idea is good, mainly considering that the number of shapes can be combined after the axis of symmetry, and the function opening is downward Considering the relationship between the axis of symmetry and -1 0 1, you can get the maximum value in a few to obtain, and that's it.
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Anonymous users2024-01-14
Your analysis is correct, this problem is unsolvable.
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Anonymous users2024-01-13
Your A is unknown, how can the maximum value not be 8,
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Anonymous users2024-01-12
Sell x 2+1 2 for the first time
Remaining x - (x 2 + 1 2) = x 2 - 1 2
The second sell(x 2-1 2) 2+1 2=x 4+1 4 leftover x 2-1 2-(x 4+1 forest number 4)=x 4-3 4 the third rotten several times sold (x 4-3 4) 2+1 2=x 8+1 8 left, the first half of the hungry spring x 4-3 4-(x 8+1 8)=x 8-7 8, and the last 0 left.
x/8-7/8=0,x=7
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