The number of real roots of equation x ^ 3 + x ^ 2 + X + a = 0 is?

The number of real roots of equation x ^ 3 + x ^ 2 + X + a = 0 is?

The cubic equation affirms three roots (including imaginary roots), but the number of real roots should be determined according to the value of A

The number of real roots of equation (1 / 2) ^ x = x ^ 1 / 2 is

(1 / 2) ^ x decreases monotonically;
X ^ 1 / 2 increases monotonically;
So: the number of real roots of equation (1 / 2) ^ x = x ^ 1 / 2 is 1;

There are five points on the plane, any three points are not in the same line, with these points as the vertex, we can draw a total of several triangles

C (5,3) = 20 triangles

What center of a triangle is equidistant from the three vertices?

The intersection of the three heights is the vertical center, the intersection of the three midlines is the center of gravity, the intersection of the three angular bisectors is the inner center, the intersection of the three vertical bisectors is the outer center, and the distance from the focus of the vertical bisector to the three vertices is equal

Which "center" of a triangle is equidistant from the three vertices?

Heart to heart

How to prove that the distance from the center of a triangle to the three sides is equal? How to prove that the distance from the center of a triangle to the three vertices is equal?

The intersection of the bisectors of the three angles of a triangle is the inner part of the triangle. According to the definition of the bisector, the distance from the point on the bisector to both sides of the angle is equal, so the distance from the inner part of the triangle to the three sides is equal

In the plane rectangular coordinate system, the vertex a (2,5), B (4,1), C (6,3) of triangle ABC
(1) If e and F are the middle points of edges AB and AC respectively, the equation of straight line EF can be solved; 2) the area of triangle can be solved
Please write down the process

(1) From the midpoint formula, we can get the e coordinate of AB midpoint is (3,3), and the f coordinate of AC midpoint is (4,4), then the slope of straight line EF k = (4-3) / (4-3) = 1. From the point oblique equation of straight line, we can get Y-3 = 1 * (x-3) y = x, which is the equation of straight line EF. (2) from the distance formula between two points, we can get: | ab | = 2 √ 5, | AC | = 2 √ 5, | BC | = 2 √ 2

∫∫[e^-(x²+y²-π)]sin(x²+y²)dxdy D:x²+y²≤π

∫∫ [e^-(x²+y²-π)]sin(x²+y²) dxdy=∫∫ [e^-(r²-π)]sin(r²) rdrdθ=e^π∫[0→2π]dθ∫[0→√π] re^(-r²)sin(r²)dr=2πe^π∫[0→√π] re^(-r²)sin(r²)d...

Let the left and right focus of the ellipse be F1, F2, and if there is a point P on the ellipse, let ∠ f1pf2 = 60, find the value range of E
I want a complete solution

When p is the vertex of the longitudinal axis, the angle is the largest, and E = sin30 ° = 1 / 2 and the eccentricity of the ellipse

F1 and F2 are the focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), P is the point on the ellipse, and the angle f1pf2 = 90 degrees. Calculate the area of the triangle f1pf2

Let f1p = m, F2P = n be defined by ellipse, M + n = 2A angle, f1pf2 = 90 degree, F1F2 = 2c, then in right triangle f1pf2, F1F2 is hypotenuse, so m ^ 2 + n ^ 2 = (2C) ^ 2, so right triangle f1pf2 area = 1 / 2 * Pf1 * PF2 = 1 / 2 * Mn = 1 / 4 * [(M + n) ^ 2 - (m ^ 2 + n ^ 2)] = 1 / 4 * (4a ^ 2-4c ^ 2) = a ^ 2-C ^ 2 = B ^ 2