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Let the first angle be x, the second angle be y, and the third angle be zx+y+z=180
x=2/3y
z-x-y=20
Add three in one formula to get z=100
Then solve the binary equation x=32 y=48
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Let the second x degree, the first 2 3x degrees, and the third x+2 3x+20 degrees, then x+2 3x+x+2 3x+20=180, the solution is x=48, so 2 3x=32 x+2 3x+20=100
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Let the second angle be x degrees, then the first angle is 2 3 x degrees, and the third angle is (x+2 3x+20) degrees, so.
x+2/3 x+x+2/3x+20=180
Just solve it.
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The degree of the third angle (180+20) 2=100°
The degree of the second angle (180-100) (1+2 3) = 48°
The degree of the first angle is 48 2 3 = 32°
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Let the first angle be a, the second angle be b, and the third angle be c, then there is a=2 3 b according to the conditions
c=a+b+20=2 3 b+b+20=5 3 b+20 according to the sum of the inner angles of the triangle is equal to 180 Yes.
a+b+c=180
Substitution. 2 3 b + b + 5 3 b + 20 = 180 to get b = 48
a=2/3 b=32
c=180-48-32=100
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The bend of the bridge mentioned in property A is generally the convex sensitive bridge boring function, which can be drawn to see.
A convex function is a function that is convex, e.g. lnx, x, 1) The title says that proof is not required, 2 is a concave function is not OK, and -x is a convex function.
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1) The function y = 2 x has property a on the defined domain d = 0, ).
Because for any x d, f(x) >f(x+1) +f(x-1)) 2, i.e., 2 x > 2 (x+1) +2 (x-1)) 2, is obviously true.
The function y = x does not have property a on the definition domain d = .
Because a set of x = 0, t = 1 can be found such that f(x) 2) lets the function y = f(x) define the domain d, and f(0) >f(1 2).
Because the function f(x) has the property a on d, for any n n, there is f(n) >f(n+1) +f(n-1)) 2.
And f(n-1) = f(n-1 + 1 2) >f(n) , so f(n) >f(n+1).
3) For the function y = g(x), the domain is defined as r and has the property a
The proposition is not true, since the key function is a strictly increasing function in the interval (- 0) does not mean that the function is also a strictly increasing function on r, we can find other functions that satisfy the property a but are strictly decreasing in other intervals.
The proposition is not true, the same goes for the above.
Hope it helps.
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Set Xiao Ming is not x this year, and his father is y this year
y+13=2(x+13)
y=3x solution gives x=13, y=39
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The father's age is x, and Xiao Ming's age is y
x=3yx+13=2(y+13)
3y+13=2y+26
3y-2y=26-13
y=13x=3y=3*13=39
The father's age is 39 years old, and Xiao Ming's age is 13 years old.
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25 minutes = 5 12 hours.
In the first case, Lingzhima did not meet:
360-100-5 12x72) (72+48)+5 12=2 and 1 3 (hours).
After the second case meets:
360+100-5 feet bucket 12x72) Mengpei (72+48)+5 12=4 (hours).
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Set: The cost of adding x vertical holes is equal 10x=4x+200 x=200 6 About 33 are about the same cost, if the slim height is less than 33, go to the store to buy in bulk to save the rise, if the addition is greater than 33, make your own savings.
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Mom gave Xiao Ming 19 apples, asked Xiao Ming to divide them into 4 piles, and asked for the first pile to be doubled, the second pile increased by one, the third pile was reduced by two, and the fourth pile was doubled, and the number of apples in these 4 piles should be the same. How do you divide the 19 apples?
Let the number of apples in the first, second, third, and fourth piles be a, according to the title: a+a=b+1=c-2=d-1 2 d=k, a+b+c+d=19
That is to say, 1 2 k+k-1+k+2+2k=19 can find k=4, so a=2, b=3, c=6, d=8
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Assuming that x wells were drilled per day, (x+3) wells were actually drilled per day.
According to the conditions, 30 x-30 (x+3)=5 simplification of the equation is obtained, and the square of x + 3x-18 = 0
Solve the equation to obtain x1 = 3 and x2 = -6 (rounded).
So it was planned to drill 3 wells a day.
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It was planned to be x per day, but now it's x+3
30/x)-【30/(x+3)】=5
Solve x=-6 (round off) or 3 (reserve).
In other words, the original plan was to drill 3 wells a day...
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It is troublesome to discuss without an equation of unknown numbers.
Drill for 5 more days, 3 more wells per day, that is, the excess part is 5 3 = 15, and the number of wells completed in the plan is 30-15 = 15
That is, when it was completed, 15 wells were completed according to the original plan, and the excess part of the original plan = 15 15 = 1, that is, the speed was doubled, so the number of wells drilled per day according to the original plan was 3 1 = 3
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The original plan was to drill x wells per day.
30/x-30/x+3=5
Simplification (x+6)(x-3)=0
Solve x=-6 or x=3, round x=-6 according to the meaning of the questionAnswer: The original plan was to drill 3 wells a day.
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1.The sum of the single digit and the ten digit is 5, the ten digit is x, then the single digit is (5-x), and the two-digit number is 9x+5 (10x+5-x).
According to the title, the sum of squares of the numbers is 17, then x 2+(5-x) 2=17
2.Let the equation be ax 2+bx+c=0
A misreads the coefficients of the x term and solves the two constants of 4 and 8 correctly, indicating that the product of the two is =c a=-32
B misreads the constant term and solves the two roots as 4 and 10 x terms, and the coefficient is correct, indicating that the sum of the two is =-b a=14
So the equation is x 2-14-32=0
3.The relationship between the two roots of the equation and the coefficients:
x1^2+x2^2=(x1+x2)^2-2x1x2=(4m-8)^2-2*4m^2=136
The solution is m=-1 or m=9
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Let the original two-digit and ten-key segments be x and the single bit be y
x+y> 10
10xy+36=10y+x
9y-9x=36
y-x=4y=x+4
2x+4>10
x>3x=4y=8
x=5y=9
The original number is 48
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Solution: The total score difference is 78-69 = 9 (points).
Let the first game be x points, the second game (x-1) points, and the third game (x-2) points.
From the inscription of Jing Yan, the year manuscript is x+(x-1)+(x-2)=9 to solve x=4 25 points per game in the volleyball game, and the average point is to continue the game after the tie.
The scores of the three games were 25:21, 25:22, and 28:26
200 (10-4) 33 pcs.
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1) CD AM CB AN CDA= ABC AC BISECTED MAN DAC= CAN=120° 2=60° AC=AC, SO ACD ACB AD=AB In rt adc, c=30° then AC=2AD and AD=AB, so AC=AD+AD=AD+AB (2) Do ce am CF an from (1) to get ace ACF then CE=CF......dac= caf=60° because e= f=90°......adc+∠cde=180° ∠adc+∠abc=180° ∴cde=∠abc……3 Ced CFB dc=bc from 1 2 3 Conclusion 1 is established AE=AC 2 in CEA, then AD=AE-DE=AC 2 - DE In the same way, AB=AF+FB=AC2 + BF is obtained from CED CFB BF=DE AD+AB=AC 2 +AC 2=AC Conclusion 2 is true, I played for half an hour, I was tired, and I did it myself.
Actually, I think so.
If you draw 5 cards of the same suit. Then you can draw one of the same as him. >>>More
exists, shifts the term to obtain: -m-2>(3-m)x, and it is easy to know that if m exists, the system of equations: >>>More
Eight years ago, eight years later, after a total of 16 years, the son was 16 years older, and if the father was 16 4 64 years older, it would still be 4 times longer, and now it is only 2 times. >>>More