(a) The first four Hermite polynomials are 1,2t,−2+4t2, and −12t+8t3. These polynomials arise naturally in the study of certain important differential equations in mathematical physics. Show that the first four Hermite polynomials form a basis of P3.(b) The first four Laguerre polynomials are 1,1−t,2−4t+t2, and 6−18t+9t2−t3. Show that these four Laguerre polynomials form a basis of P3.

first anwser choice.

step-by-step explanation:

the greater/less or equal to symbol (> or < with the line at the bottom) is a closed/filled in circle.

-5c + 2 _< _27

subtract from both sides by 2

-5c _< _ 27

divide both sides by -5

c _< _ 5.4

flip the inequality sign

c _> _5.4

the only possible answer that would match the simplified inequality is the first answer choice

point a is the center of the circle that passes through points e, f, and g and the center of the circle that passes through points x, y, and z.

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!