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1.Take a point on one of the lines and find the distance from the other.

Or, if you draw a diagram, the y-axis intercept is -1 and 1 2, using the Pythagorean theorem, the hypotenuse is 3 2, and the slope is 3 2, then the distance between the two straight lines is (3 2) * (2 root number 13).

A 2-b 2=c 2, and a = 2c then there is b 2=3c 2=4 2a 2=64 3

Standard equation: x 2 64 3 + y 2 16 = 13Derivatives can be found Curve y=inx Derivative: y=1 x Curve y=x2 Derivative: y=2x

The slopes are equal to give 2a 2=1

1.Line 3x-2y+1=0 crossing point (1,1).

The distance between this point is 3x-2y-2=0 |3-2-2|/√(3^2+2^2)=√13/13

a=2c a^2=b^2+c^2

Get a 2 = 64 3

So the standard equation is 3x 2 64 + y 2 16 = 1 derivative y'=1/x

y=x2 derivative y'=2x

The tangents at the point x=a are parallel to each other, i.e., the derivatives are equal: 1 x=2xx=a= 2, 2

These are basic questions, so it is recommended that you review them more.

It's so hard to take a screenshot, forget it, only the last question.

Approximate with Taylor series, replace (.

First, find the intersection points as (-2,5) and (2,-3).

Then the area = (2 to 2) [(2x+1)-(x 2-2x-3)] dx (because the function value of the point on the line in the range is definitely greater than that of the parabola, so each microelement is positive, regardless of whether it is above or below the x-axis, their difference is always positive).

-2 to 2) [(2x+1)-(x 2-2x-3)] dx = (2 to 2) (4-x )dx

-2 to 2) (4x-x 3).

A very simple double integral, this graph is both x-type and y-type... Directly take the value range of x at the intersection of two lines, and the value range of y is OK·· Uh·· I'll give you a formula... Standard column: d = 2 to 2) [ x 2-2x-3 to -2x+1)xydy]dx··· If you don't know how to double integrate, when I didn't say... Huh...

If you use double integrals to determine which quadrant you are in... It's all the same...

I thought that a parabola intersected a straight line, and it should be calculated using integrals.

The derivation process is as follows:

y=x^2-2x-3=f(x)

y=-2x+1=g(x)

Get x 2-2x-3=-2x+1

x^2=4x=2 or x=-2

The area is (g(x)-f(x))dx xe(-2,2)= 2x+1-x 2+2x+3)dx= x 2+4)dx=(-1 3x 3+4x) xe(-2,2).

Pair -2x+1-x 2+2x+3=-x 2+4

In the above relation from -2 to 2 integrals, (-1 3)*2 3+8-8 3+8=16-16 3=32 3

Find the integral of the difference between the 2nd order function and the intersection point of the x-axis and 0 between the 2nd degree y 0 and the 2nd order function.

An indefinite system of equations with an infinite number of solutions.

Several sets of solutions are given, x=, y=, z=0

x=0，y=，z=

x=，y=0，z=