How to calculate X - (1 / 6 + 3 / 8) = 1 / 4

How to calculate X - (1 / 6 + 3 / 8) = 1 / 4

X - (1 / 6 + 3 / 8) = 1 / 4
X-24 13 / 4 = 1 / 4
X = 1 / 4 + 13 / 24
X = 19 / 24

The problem is 8.5-x = 3.7, X divided by 1.2 = 0.8, 25X = 17

8.5-X=3.7
X=8.5-3.7
X=4.8
X divided by 1.2 = 0.8
X=0.8*1.2
X=0.96
25X=17.5
X=17.5/25
X=0.7

Find four numbers 1,3,5,13 and the result is equal to 24

(13-1)×(5-3)=24;
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In rectangular paper ABCD, ab = 3cm, BC = 4cm. Now fold and flatten the paper so that a and C coincide. If the crease is EF, the area of the overlapping part △ AEF is equal to______ ．

Let AE = x, from the folding, EC = x, be = 4-x, in RT △ Abe, AB2 + be2 = AE2, that is 32 + (4-x) 2 = X2, the solution is: x = 258, from the folding, we can know ∠ AEF = ∠ CEF, ∵ ad ‖ BC, ∵ CEF = ∠ AFE, ∵ AEF = ∠ AFE, that is AE = AF = 258, ∵ s △ AEF = 12 × AF × AB = 12 × 258 × 3 = 7516

If f (x) = x Λ 2 + AlN (x + 1), where a ≠ 0.a ＜ 1 / 2, find the extremum of F (x)

Domain x > - 1
F (x) derivative = 2x + A / (x + 1) = (2x Λ 2 + 2x + a) / (x + 1)
Because discriminant = 4-8a > 0
F (x) derivative = 0 to get x = (- 1-radical (1-2a)) or x = (- 1 + radical (1-2a))
X = (- 1-radical (1-2a)) - 1
So the extreme point (- 1 + radical (1-2a)) of function f (x)

How much is output - 3 - 2 4?
Enter x + 2x △ (- x) + 2

0,1,1,1

X + Y-2 is a factor of bivariate quadratic x ^ 2 + axis + by ^ 2-5x + y + 6, then a + B=

Let another factor be a
x^2+axy+by^2-5x+y+6=A(x+y-2)
When x = 1, y = 1, the right side = a * (1 + 1-2) = 0
Then the left side is also equal to 0
x=1,y=1
x^2+axy+by^2-5x+y+6=0
So 1 + A + B-5 + 1 + 6 = 0
a+b=-3

In the cube abcd-a1b1c1d1 with edge length 1, the distance between plane ab1c and plane a1c1d1 is? Can coordinate method be used?

OK. In addition, the title you typed is wrong. It should be plane ab1c and plane a1c1d
Let's build A1 (0,0,0) B1 (1,0,0) a (0,0,1), D1 (0,1,0) coordinate system
The geometric center of C (1,1,1) △ ab1c is p: 1 / 3 (a + B1 + C) = 1 / 3 [(0,0,1) + (1,0,0) + (1,1,1)] = (2 / 3,1 / 3,2 / 3)
∵B=(1,0,1) ∴PB=√[ (1-2/3)² + (1/3-0)² + (1-2/3)²] = √3 / 3
∵ C1 = (1,1,0) diagonal length of cube AC1 = √ (1 + 1 + 1) = √ 3
The distance between plane ab1c and plane a1c1d1 is ac1-2pb = √ 3 / 3

What is the interval of F (x) = the x power of E + the zero point of X-2?

If f (x) = e ^ x + X-2, then f (0) = - 10, the interval of zero point is (0,1)

As shown in the figure, Pb is the projection of oblique line PA of plane a in plane A. let angle APB = angle 1, angle BPC = angle 2, and angle APC = angle 3?

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