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1. Let the residual amount be y, then, y=10t - 24 (5t) +100[ 10t)] 2 - 2* 10t) *6 2) +6 2) 2 -(6 2) 2 +100

10t) -6√2]^2 - 72+100[√10t) -6√2]^2 +28

When t = hour, the minimum value of y is 28

2. Y < 30 then [ 10t) -6 2] 2 +28<30, that is, [ 10t) -6 2] 2 < 2

then - 2< 10t) -6 2 < 25 2 < 10t) <7 2

50< 10t< 98

5. That is, less than 30 tons of materials per hour.

1. Let the residual quantity be y, then, y=100+10t - 24 (5t)[10t)] 2 - 2* 10t) *6 2 +(6 2) 2 +28

10t) -6 2] 2 +28, 0 t 24(10t) =6 2, that is, 10t=72, that is, when t = y takes the minimum value of 282, y < 30 then [ 10t) -6 2] 2 +28<30, that is, [ 10t) -6 2] 2 < 2

then - 2< 10t) -6 2 < 25 2 < 10t) <7 2

50< 10t< 98

5. That is, less than 30 tons of materials per hour.

The remaining amount of material in the 7th hour is the smallest, and the remaining amount is tons.

In 24 hours, there are 4 hours when the remaining material volume is less than 30 tons.

If you can't see clearly, you won't type.

c……The sine theorem knows no: sina a = sinb b b = sinc c

So a:b:c=5:11:13

Then the Pythagorean theorem can be determined.

sina:sinb:sinc=5k:11k:13k large side to large angle.

cosc=5k×5k+11k×11k-13k×13k÷5k×11k×2＜0

So angle c is an obtuse angle.

c must be an obtuse triangle.

With the sine theorem.

a sina = b sinb sinc = 2r (r is the radius of the triangle abc circumscribed circle).

i.e. a:b:c=sina:sinb:sinc=5:11:13, so a 2 + b 2 so choose c

According to the sine theorem a:b:c=sina:sinb:sinc=5:11:1313 2=169>5 2+11 2=146, so it is an obtuse triangle c

Don't need a master, I'm a low-handed hand to tell you, you should choose C

From the sinusoidal theorem and the three sides of abc a:b:c=5:11:13, it is an obtuse triangle (5 + 11 13, so it is an obtuse triangle).

The sine theorem, you can set the three sides to 5x, 11x, 13x, and use the cosine theorem to calculate the opposite angle of 13x, there is cosa<0, which is an obtuse triangle.

c 5 squared plus 11 squared 13 squared.

by sina:sinb:sinc=5:11:13

It can be seen that the three-sided length ratio is 5:11:13 (the name of the theorem was forgotten.) ）

5:12:13 is a right triangle, so this one is an obtuse triangle.

The question ab is on a circle with o as the origin and 2 as the radius.

i.e. |Vector oa|=|Vector ob|=√2

x1x2+y1y2=vector oa*vector ob=|oa||ob|cos=√2x√2cos150°=-1

x1x2+y1y2 is the oa vector dot multiplied ob vector and because the angle is 150 degrees.

So x1x2+y1y2 =|oa||ob|cos(150°)|oa|=|ob|=r=root number 2

So the result is -root number 3

cos2a=-3 5, and cos2a=2cos a-1, so 2cos a-1=-3 5, cos a=1 5, because a is the third quadrant angle, so cosa<0, cosa=- 5 5

and sina<0, sina=-2 5 5

tana=sina/cosa=2, tan2a=2 tana/(1 – tan²a)=-4/3.

tan（π/4+2a)=( tanπ/4+ tan2a)/(1- tanπ/4tan2a)=-1/7.

In the third quadrant, the tan angle is greater than 0

So tan2a=4 3

tan（π/4+2a)=(tanπ/4+tan2a)/(1-tanπ/4tan2a)

The proportional number is a2a4=(a3) 2

That is, (a3) 2=1, and an>0, so a3=1 sets the common ratio to q, a2=a3 q=1 q, a4=a3q=qa1=a3 q 2=1 q 2

s3=a1+a2+a3=7

1/q^2+1/q+1=7

1/q^2+1/q-6

1/q+3)(1/q-2)=0

q>0, then we get 1 q=2, q=1 2

a1=1/q^2=4

So, s5 = a1 * (1-q 5) (1-q) = 4 * (1-1 2 5) (1-1 2) = 8 (1-1 32) = 31 4

Solution: Let q be the common ratio of the series. Because a2a4=1 i.e. a2a2q 2=1, a2=1 q

And because a3=a2q, a3=1And because a2=a1qSo a1=1 q 2

According to the title, s3=7 is a1+a2+a3=7 i.e. 1 q 2 + 1 q + 1 = 7Solution: q=1 2 So, a1=4

So, s5=a1(1-q 5) (1-q)=4[1-(1 2) 5] (1-1 2)=31 4

Answer: So s5 is equal to 31 4

The key to this problem is to find the breaking point, a2a4=a1q multiplied by a1q 3=a1q 2 times a1q 2=a3 squared = 1 is a positive proportional series.

So a1>1 q>0 so a3=1 s3=a1+a2+a3=7=a3 q 2+a3 q+a3

That is, there is 6q 2-q-1=0 q>0 solution q= a3=1 so: a1=4 a2=2 a3=1 a4= a5= so s5= solution.

Because f(a) = f(b).

Substituting the analytic formula to get |lga|=|lgb|

So LGA = LGB

Because 01, a+2b=a+2 a

Let g(a)=a+2 a (0g.)'(a)=1-2 a<0 is constant, so g(a) is a subtraction function.

g(1)=3, but a cannot be 1, so the minimum value of 3 cannot be taken.

Therefore, the value range is (3,+).

By the title |lga|=|lgb|

lga=±lgb

0＜a＜b，∴lga≠lgb

LGA=-LGB, AB=1 (where 0 A 1 B) B=1 A makes y=A+2B=A+2 A

This function is subtractive on (0, 2), and the range of a is (0,1)y=a+2b=a+2 a is subtractive on (0,1). y＞3

y=|lgx|, to make its image, and because f(a) = f(b) and 0

There's no maximum in the first place, right?

It's a matter of finding the minimum value.

Then set the first floor of g(a) a+2 a(0

The graph of this function should be very clear, so I won't draw it if it's inconvenient.

After drawing the picture, it is clear that the upper limit of this value range must be infinite, but there is a lower limit.

Let f(a)=f(b), then | lga |=| lgb |, because 01 so a+2b = a+2 a

Let g(a)=a+2 a (03

Therefore, the value range is (3,+).