Who s going to help? There are a lot of questions that I can t do 30

2 (1) Unclear Question Tell you how.

Step 1: From 2 unequal real roots, we can know that the square of b is -4ac>0 (this inequality can find a range of m)Step 2: Let the real roots of 2 be a b to know that 1 a+1 b is greater than or equal to -2 and further obtain (a+b) ab is greater than or equal to -2 and use the root finding formula to know a+b=?

ab=?The above two inequalities can be found in common.

2) Because a:b=1:4, b=4a has a known a+b=?Just go in on your behalf.

=1-4a≥0

a≤1/4x1=[-1+√(1-4a)]/2, x2=[-1-√(1-4a)]/2

Absolute value a + absolute value b

Absolute value [-1+ (1-4a)] 2 + absolute value [-1- (1-4a)] 2

If 0 a 1 4, the absolute value a + the absolute value b

Absolute value [-1+ (1-4a)] 2 + absolute value [-1- (1-4a)] 2

1-√(1-4a)]/2+[1+√(1-4a)]/2

If a<0

Absolute value [-1+ (1-4a)] 2 + absolute value [-1- (1-4a)] 2

(1-4a)-1]/2+[1+√(1-4a)]/2

1-4a)m≠0=(8m+1)^2-64m^2=16m+1>0

m>-1/16

x1+x2=-(8m+1)/4m^2,x1*x2=1/m^2

1/x1+1/x2=(x1+x2)/(x1*x2)=-(8m+1)/4≥-2

The value range of m 7 8m is -1 16x1+4x1=-(8m+1) 4m 2,x1*4x1=1 m 2

x1*4x1=1/m^2

x1 = 1 2m or x1 = -1 2m

x1+4x1=-(8m+1)/4m^2

5x1=-(8m+1)/4m^2

x1=1/2m

5/2m=-(8m+1)/4m^2,m=-1/18

Or. x1=-1/2m

5/2m=-(8m+1)/4m^2,m=1/2

x1*x2=a absolute value a+absolute value b=(a>=0),1; (a<0), root numbers 1-4a

Draw the gourd in the same way.

Start by calculating a=3

Bring it in again. Get the answer ==

Solution: When making a circular motion in the vertical direction, the water in the bucket should not flow out at the highest point in the vertical plane, mg ten n = ma, completely take mg as the centripetal force, at this time n = o so that a is the smallest, and the obtained velocity is the smallest a = g This is the acceleration of the circular motion mg two ma two mv 2min r vmin= gr= m second, (2) when v = 3m s m second, mg ten n two mv2 r n n=

At the highest point, the water does not overflow.

Gravity fully provides centripetal force = centrifugal force.

The second question is, velocity produces centrifugal force, gravity - centrifugal force = pressure.

That's how it should be.

Solution: At the highest point, the water does not flow out, which means that only when the water is completely weightless, it will not flow out, that is, mg ten n = ma, when a = g n = 0, at this time the acceleration of circular motion is g, if g two 10 m s 2 take the water in the bucket as the research object: ma = mg two mv 2min r vmin= gr two 6 two meters second (2) mg ten n two mv 2 r, substitute the data to obtain n two one mg + mv 2 r = cow.

Featuring x candies.

3a+2=x

5b+3=x

7c+2=x

So, 3a = 7c

The solution is x=23 (minimum).

Then this number is subtracted by 2, it is a common multiple of 3, 7, you can let this number be 21k+2 (k>=1), 21k+2) divided by 5 remainder is equivalent to k+2 divided by 5 remainder, if it is 3, then k is the minimum 1, this number is the minimum is 23

15/3=5, 24/4=6, 27/5=;

If the proportions are met, the numbers should be equal, obviously: "5, 6, ; The smallest is 5 (fruit candy), so the fruit candy runs out first.

If you want to mix it all with assorted sugar, you have to buy as much sugar as possible.

The most sugar is toffee, and its ratio is: 6, so 3*6-15=3 5*6-27=3

Buy 3 kg of fruit candy and 3 kg of chocolate candy.

If 1 serving of assorted candy requires 3 kg of fruit candy, 4 kg of toffee, and 5 kg of chocolate candy, then 15 kg of fruit candy can make 15 3 = 5 parts of assorted candy, 4 * 5 = 20 kg of toffee, and 5 * 5 = 25 kg of chocolate candy. So when the fruit candy runs out, the milk candy and chocolate candy are left, and the fruit candy runs out first.

24 kg of toffee can make 24 4 = 6 servings of assorted sugar, you need 3 * 6 = 18 kg of fruit candy, and 5 * 6 = 30 kg of chocolate candy.

It is necessary to buy 18-15 = 3 kg of fruit candy and 30-27 = 3 kg of chocolate candy.

to be able to mix them all into assorted candies.