Given the equations x + y + Z = 5,2x + 3Y + Z = 8,3x + 4y-z = 4, the method of finding an unknown number at one time is a, 1 + 2 + 3, B, 1 + 3-2, C1 + 2-3, D, 2 + 3-1,

Given the equations x + y + Z = 5,2x + 3Y + Z = 8,3x + 4y-z = 4, the method of finding an unknown number at one time is a, 1 + 2 + 3, B, 1 + 3-2, C1 + 2-3, D, 2 + 3-1,

a: 6X + 8y + Z = 17
b: 2X + 2y-z = 1 cannot
c：3z=9 z=3 ∴x+y=2 2x+3y=5 y=1 x=1
d: 4X + 6y-z = 7 cannot

In the equations 2x + y = 1-m, 3x + 2Y = 2 + m, if the unknowns X and y satisfy x + Y > 0, then the value range of M is?

From 2x + y = 1-m, 3x + 2Y = 2 + m we can get
3x+2y-(2x+y)=2+m-(1-m)
x+y=2m+1
From x + Y > 0, we can get x + y = 2m + 1 > 0, so m > - 1 / 2

3x + 6x = (3.6 + 2.4) × 3 / 10

3x + 6x = (3.6 + 2.4) × 3 / 10
9x = 6 × 3 / 10
9x = 9 / 5
X = 9 / 5 △ 9
X = 1 / 5
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In the triangle ABC, the angle BAC = 24 degrees, ad bisects the angle BAC, crosses a to make the vertical line BC of DA at point m, if BM = Ba + AC, calculate the degree of the angle ABC
This is the second day of the problem, do not use the function Oh!

Extend B to AE to make AE = AC, connect me, extend Ma to CE to F
AC=AE AEC=ACE=1/2BAC=CAD
Ad / / CE Ma vertical ad Ma vertical EC
AE=AC EF=CF
MF vertical CE EM = MC MEC = MCE
AEC=ACE MEB=ACB
BM=AB+AC=BE BME=BEM
MEC+MCE+BEM=180
3EMB+2AEC=180 2AEC=BAC=24
EMB=56
ACB=MEB=EMB=56

Solution equation: (2x + 3) (x-4) - (x + 2) (x-3) = xx-8

(2x+3)(x-4)-(x+2)(x-3)=xx-8
2x^2-5x-12-(x^2-x-6)=x^2-8
2x^2-5x-12-x^2+x+6=x^2-8
x^2-4x-6=x^2-8
-4x-6=-8
-4x=-2
x=1/2

Ellipse x ^ 2 / 4 + y ^ 2 = 1, intersection of line L passing through the right focus of ellipse, ellipse and two points a and B, make a circle with diameter AB passing through the point, and find the equation of line L

A ^ 2 = 4, B ^ 2 = 1, so C ^ 2 = a ^ 2-B ^ 2 = 3, the right focus of ellipse is (√ 3,0), let the equation of line l be y = K (x - √ 3), substitute it into the elliptic equation to get x ^ 2 / 4 + K ^ 2 (x - √ 3) ^ 2 = 1, simplify to (4K ^ 2 + 1) x ^ 2-8 √ 3K ^ 2 * x + 12K ^ 2-4 = 0, let a (x1, Y1), B (X2, Y2), then X1 + x2 = 8 √ 3K

As shown in the figure, in △ ABC, CD ⊥ AB, foot drop is D, point E is on BC, EF ⊥ AB, foot drop is F. (1) is CD parallel to ef? Why? (2) If ∠ 1 = 2 and ∠ 3 = 65 °, then ∠ ACB=______ Degree

(1) ∵ CD ⊥ AB, perpendicular foot D, point E on BC, EF ⊥ AB, ∵ CD ∥ ef (two lines perpendicular to the same line in the plane are parallel to each other); (2) ∵ EF ∥ DC ∥ 2 = ∠ BCD (two lines are parallel and the same angle is equal) ∵ 1 = ∠ 2 (known) ∵ 1 = ∠ BCD (equivalent substitution) ∵ DG ∥ BC (internal error)

Let Z ∈ C be a pure imaginary number zz − 1, and find the maximum value of | Z + I |

Let z = x + Yi, x, y ∈ R, since zz-1 = x + yix-1 + Yi = (x + Yi) (x-1-yi) (x-1 + Yi) (x-1-yi) = x2 + y2-x (x-1) 2 + Y2 + y (x-1) 2 + y2i is a pure imaginary number, there is X2 + y2-x = 0y ≠ 0, that is, (X-12) 2 + y2 = 14 & nbsp; (Y ≠ 0), which means a circle with C (12,0) as the center and R = 12 as the radius

Given 2x = log2 3, find the value of (2 ^ 3x-2 ^ - 3x) / (2 ^ X-2 ^ - x)

The moon is purple,
2X＝log2(3)
X＝1/2log2(3)
2^x＝2^log2(√3)
2^x＝√3
2^(3*x)－2^(－3x)/(2^x－2^－x)
＝(2^x－2^(－x))(2^2x＋1＋2^(－2x))/(2^x－2^(－x))(2^x＋2^(－x))
＝(2^2x＋1＋2^(－2x))/(2^x＋2^(－x))
＝((2^x＋2^(－x))^2－1)/(2^x＋2^(－x))
＝(2^x＋2^(－x))－1/(2^x＋2^(－x))
＝(√3＋1/√3)－1/(√3＋1/√3)
＝(3＋1/√3)－(1/(3＋1)/√3)
＝4/√3－(√3/4)
＝4√3/3－(√3/4)
＝13√3/12

The radius of the earth is 6370 km, and the radius of table tennis is about 2 cm, so the radius of table tennis is twice that of the earth______ One third

The radius of the earth is 6370 km = 637 million cm, the radius of the earth is 2 △ 637 million = 1318 million