x+2y=1,3x-2y=11

x+2y=1,3x-2y=11

Add the two formulas to get 4x = 12, x = 3, and then bring x = 3 into x + 2Y = 1 to get y = - 1, so x = 3, y = - 1

How to change (x-3) ^ 2 / 25 + y ^ 2 / 16 = 1 into a polar equation
It turns into 25p ^ 2-9p ^ 2 (COS α) ^ 2-96pcos α = 256, right?
How to make 25p ^ 2-30pcos α + 9p ^ 2 (COS) ^ 2 = 256

Dizziness
It's using
x=pcosa,y=psina
(x-3)^2/25+y^2/16=1
16(x-3)^2+25y^2=400
And then just unfold

What is the polar equation of x = 2

pcosθ=2

If a point a (1,2) on y = 2x ^ 3 is known, then the tangent slope of point a is equal to?

The derivative function of y to X is y '(x) = (2x ^ 3)' = 6x ^ 2, so the slope at point a (1,2) is k = y '(1) = 6 * (1 ^ 2) = 6

There is one thing I don't understand about the concept of the slope of the tangent of a curve by derivative
When the point PN (xn, f (xn)) approaches the point P (x0, f (x0)) along the curve y = f (x), why are there two slopes, one Secant and one tangent, not all the same?

Is the derivative of a curve the slope of its tangent

The derivative of a curve is a function of the slope of the tangent of the curve

Let f (x) = 2 ∥ X-1 ∥ + ∥ x + 2 ∥ replace the sign of absolute value! Find the inequality F（
Let f (x) = 2 ∥ X-1 ∥ + ∥ x + 2 ∥ replace the sign of absolute value! Find the solution set f (x) of inequality f (x) ≥ 4

1、 1. When x = x > = 0 and x > = 4 / 3 union
2、 1. When x = 6;
2. When - 2

Is y = I 2x * x I-3 a functional relation? Two sides of 2x * x are absolute value symbols. The teacher did not give a clear answer
My puzzle is: when x takes a pair of opposite numbers, y is the same corresponding value
For example: x = 1 and x = negative 1, y is equal to - 1
Can a function value correspond to the values of two independent variables at the same time?

As long as X does not correspond to two or more Y values, a function value can correspond to the values of n independent variables at the same time
Let's say the constant function y = 1, which means that when x takes all values, y takes 1

What is the necessary and sufficient condition for arccos (- x) to be greater than arccos x?

arccosx∈[0,∏]
(1) When arccos x ∈ [0, Π / 2], arccos (- x) ∈ (Π / 2, Π] > arccos X
(2) When arccos x ∈ (Π / 2, Π), arccos (- x) ∈ [0, Π / 2]

If arccos (- x) < arccos x holds, the value range of X is
It's a process

Because arccosx ∈ [0, Π]
(1) When arccos x ∈ [0, Π / 2], arccos (- x) ∈ (Π / 2, Π] > arccos X
(2) When arccos x ∈ (Π / 2, Π), arccos (- x) ∈ [0, Π / 2]