What is the maximum area of an inscribed rectangle in an ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1?

What is the maximum area of an inscribed rectangle in an ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1?

∵ the ellipse is symmetric about the origin and the coordinate axis, and the inscribed rectangle is also symmetric about the origin and the coordinate axis
According to the elliptic equation, the point of the rectangle in the first quadrant can be set as: (2cos α, √ 3sin α), 0

300. OK, 30
I've got a question,

3X+5X=48 14X-8X=12 6*5+2X=44 20X-50=50 28+6X=88 32-22X=10 24-3X=3 10X*（5+1）=60 99X=100-X X+3=18 X-6=12 56-2X=20 4y+2=6 x+32=76 3x+6=18 16+8x=40 2x-8=8 4x-3*9=29 8x-3x=105 x-6*5=42 x+5=7 2x+3=10 12x-...

Given the function f (x) = (2a + 1 / a) - (1 / A & # 178; x), the constant a > 0
1. Let Mn > 0 and m < n, prove that f (x) increases monotonically on [M, n]
2. Let 0 ＜ m ＜ N and f (x) have both the domain of definition and the domain of value [M, n], and find the maximum value of N-M

1. According to the meaning of the topic, the domain of definition is in (- ∞, 0) (0, + ∞), and Mn > 0. It can be seen that m and N belong to (- ∞, 0) (0, + ∞). F (m) - f (n) = (2a + 1 / a) - (1 / A & # 178; m) - [(2a + 1 / a) - (1 / A & # 178; n)] = 1 / A & # 178; n-1 / A & # 178; m = 1 / A & # 178; (1 / n-1 / M) = 1 / A & # 178; * (M -...)

Let π / 6 be a zero point of function f (x) = sin (2x + Φ), then the sum of all extreme points of function f (x) in the interval (0,2 π) is____ Thank you for your detailed answer. The answer is 14 / 3 π

Let π / 6 be a function
If f (x) = a zero point of sin (2x + Φ), then the sum of all extreme points of function f (x) in the interval (0,2 π) is____
Analysis: ∵ π / 6 is a zero point of the function f (x) = sin (2x + Φ)
∴f(π/6)=sin(2π/6+Φ)=0==>2π/6+Φ=0==>Φ=-π/3；
f(x)=sin(2x-π/3)==>f’(x)=2cos(2x-π/3)=0==>2x-π/3=2kπ+π/2==>x=kπ+5π/12；2x-π/3=2kπ+3π/2==>x=kπ+11π/12
∵ on the interval (0,2 π)
5π/12+17π/12+11π/12+23π/12=(5+17+11+23)π/12=14π/3

Mathematics Grade 5 Volume 1 fraction calculation and fraction equation

Tangent equation of curve y = x ^ 3-5X + 1 at point (2, - 1)? How to get it? Please answer it in detail
We know that the slope is 7, so we can get the equation Y-1 = 7 (X-2)
I don't know how to do this

Let me see, the curve equation is y = x ^ 3-5X + 1, so its derivative is y '= 3x ^ 2-5. Substituting the value of X in (2, - 1), that is, 2, into y', we can get y '= 12-5 = 7, so we can know that the slope is 7. Now we know the slope, and a point on the tangent (2, - 1), we can bring it into the tangent equation. Finally, we can get the tangent equation y + 1 = 7 (X-2), and simplify it to get the tangent equation y-7x + 15 = 0, which is detailed enough and despises the copying masters

It is known that in △ ABC, ab ≠ AC, the proof is: ∠ B ≠ C
A. ∠A=∠BB. AB=BCC. ∠B=∠CD. ∠A=∠C

The opposite side of ∠ B ≠ C is ∠ B = ∠ C. Therefore, it can be assumed that ∠ B = ∠ C. Therefore, C is selected

Formula for finding the final term of arithmetic sequence

Let the first term be A1 and the tolerance be D, then the last term be an = a1 + (n-1) d

6(2/1x-4)+2x=7-(3/1x-1)

3x-24+2x=7-1/3x+1
3/16x=32
x=6

On a square cardboard with a circumference of 4cm, cut the largest circle. What is the area of the remaining part in square centimeter?

4 △ 4 = 1 (CM), 1 × 1-3.14 × (1 △ 2) 2, = 1-3.14 × 0.25, = 1-0.785, = 0.215 (square cm), a: the area of the remaining part is 0.215 square cm