To solve the equation, X is 8 of 9 = 6.4.5x of 5 + x = 11.5 of 17-5x of 6 = 1 of 3

To solve the equation, X is 8 of 9 = 6.4.5x of 5 + x = 11.5 of 17-5x of 6 = 1 of 3

（8/9）X=6/5
X=6/5*9/8
X=27/20
2.4.5X+X=11/17
5.5X=11/17
X=2/17
3.5-5/6X=1/3
5/6X=14/3
X=28/5
Solve the equation 3.5x-x + 8 = 12
3.5X-X+8=12
2.5X+8=12
2.5X=12-8
2.5X=4
X = 4 divided by 2.5
X=1.6
x=1.6
2.5x=12-8
x=1.6
It's too easy
Solve the equation. 5x = 60 x-43 = 17 16 + x = 28 x-14.8 = 4
solve equations.
5x=60 x-43=17 16+x=28
x-14.8=4.2 x÷1.4=5 x+9=24
x＝60÷5
=
x＝17+43
=
x＝28-16
=
5x-8 * 3 = 24 solution equation
5x-8*3=24
5x-24=24
5x=48
x=48/5
5x-8*3=24
5x-24=24
5x=48
x=9.6
5x-8*3=24
5x-24=24
5x=48
x=9.6
5x-8*3=24
5x=48
x=9.6
5x-8*3=24
5x-24=24
5x=48
x=9.6
Given x2 + xy = 3, Y2 + xy = - 2, then what is 2x2-xy-3y2
2x²-xy-3y²
=2(x²+xy)-3(xy+y²)
=2×3-3×(-2)
=12
x2+xy=3 (1)
y2+xy=-2 (2)
(1)×2-(2)×3
×2x2-xy-3y2=6-6=0
The result is 12
If the square of quadratic function y = ax - 4x + 1 has the minimum value - 1, then the value of a is
If there is a minimum, the opening of quadratic function is upward, a > 0
Y=AX²-4X+1=A(X-2/A)^2+1-4/A^2>=1-4/A^2
The minimum value is - 1
1-4/A^2=-1
4/A^2=2
A^2=2
A = root 2
If you don't understand, you are welcome to ask,
Just find out when the function takes the minimum. Since there is a minimum value, the opening of the function is upward, and the axis of symmetry is x = 2 / A. the minimum value is obtained at the axis of symmetry. Substituting x = 2 / a into the function, y = 4 / A-8 / A + 1 = - 1, then - 4 / a = - 2, a = 2
The sum of the numbers in each digit of a three digit number is equal to 12, and its one digit number is 2 less than the ten digit number. If its hundred digit number and one digit number are interchanged, the result is obtained
The number is 99% smaller than the original number
Let's assume that the 100 and 10 digits are ABC
c=b-2
The original three digits are 100A + 10B + B-2 = 100A + 11b-2
The three digit number is 100 (b-2) + 10B + a = 110b + A-200
New number = original number - 99
Then 100A + 11b-101 = 110b + A-200
99a+99=99b
a+1=b
a=b-1
a+b+c=12
So B-1 + B + B-2 = 3b-3 = 12
B=5
A=4
C=3
The original number was 453
I hope the answer will be useful to you
100a+10b+c
a+b+c=12
b-2=c
(100c+10b+a)+99=100a+10b+c
A=4
B=5
C=3
Four hundred and fifty-three
The number of hundreds is 99 / (100-1) = 1
Single digit (12-2-1) / 3 = 3
Ten digit 3 + 2 = 5
Hundred digit 3 + 1 = 4
The original three digit number is 453
Let X be ten, then X-2 be one and y be one hundred
2x+y-2=12
100(x-2)+10x+y+99=100y+10x+x-2
The solution is: x = 5, y = 4, X-2 = 3
The original number is 453
If the individual digit is a, then the ten digit is (a + 2) and the hundred digit is (10-2a).
From the meaning of the title: 100 (10-2a) + A-100A - (10-2a) = 99
99﹙10－2a﹚－99a＝99
10－2a－a＝1
3... Unfold
If the individual digit is a, then the ten digit is (a + 2) and the hundred digit is (10-2a).
From the meaning of the title: 100 (10-2a) + A-100A - (10-2a) = 99
99﹙10－2a﹚－99a＝99
10－2a－a＝1
3a＝9
a＝3
So, this three digit number is 453
Four hundred and fifty-three
Y set ten digits as X and hundred digits as y
100Y+10X+(X-2)=100(X-2)+10X+Y+99
X-Y=1
The number of tens is 1 larger than that of hundreds, and the sum is 12 (12-1-2) / 3 = 3, so 3 + 1 = 4, 3 + 2 = 5
The original number is 453
Let the hundreds of the three digits be a, the tens be B, and the ones be c
According to the sum of the numbers on each digit is equal to 12, the equation a + B + C = 12 ① is obtained
According to the fact that its one digit number is 2 less than ten digit number, B = C + 2
Substituting ① a + C + 2 + C = 12
A + 2C = 10 ③, put it first
According to the fact that the number obtained is 99 less than the original number, the equation 100A + 10B + C - (100C + 10B + a) = 99 is obtained
Remove bracket: 100A + 10B +... Expand
Let the hundreds of the three digits be a, the tens be B, and the ones be c
According to the sum of the numbers on each digit is equal to 12, the equation a + B + C = 12 ① is obtained
According to the fact that its one digit number is 2 less than ten digit number, B = C + 2
Substituting ① a + C + 2 + C = 12
A + 2C = 10 ③, put it first
According to the fact that the number obtained is 99 less than the original number, the equation 100A + 10B + C - (100C + 10B + a) = 99 is obtained
Remove brackets: 100A + 10B + c-100c-10b-a = 99
99a-99c=99
99(a-c)=99
a-c=1④
The system of quadratic equations with two variables is obtained
a+2c=10③
a-c=1④
③ - ④ = 3C = 21, C = 3, so a = 4, B = 5
So the original number is 453
I'm glad to solve the problem for you, hope to adopt it. If you don't know, please ask me
The correct writing and solution of the algebraic expression Y2 + (x2 + 2xy-3y2) - (2x2-xy-2y2)
2y+(2x+2xy-3*2y)-(2*2x-xy-2*2y)
2y+2x+2xy-3*2y-2*2x+xy+2*2y
2y+2x+2xy-6y-4x+xy+4y
2y-6y+4y+2x-4x+2xy+xy
-2x+3xy
If the minimum value of quadratic function y = x square - 4x + n is 3 n=
y=x²-4x+n
=x²-4x+4+n-4
=(x-2)²+n-4
When x = 2, the minimum value n-4 is obtained
n-4=3
N=7
For a 3-digit number, the 100 digit number is one more than the 10 digit number, and the single digit number is three times less than the 10 digit number by two. The sum of this number and its reciprocal is equal to 1171?
There is a problem with the problem!
The original question is: the number on the hundred digit is one times larger than that on the ten digit, and the number on the one digit is three times less than that on the ten digit. He reversed the hundred digit and the one digit, and the sum of the three digit and the original three digit is 1171
answer
Let x, y and Z be the number of three digits, then:
z=y+1 …… (1)
x=3y-2 …… (2)
(100z+10y+x)+(100x+10y+z)=1171 …… (3)
By introducing the two expressions (1) (2) into (3), we get the following result:
(100y+100+10y+3y-2)+(300y-200+10y+y+1)=1171
The solution is: y = 3, so: x = 7, z = 4
This three digit number is 437
Four hundred and thirty-seven