Let A be the set of all polynomials in Q [x] with zero constant term. Prove this set is an ideal in q[x].

i think it is everything has to equal 180 so, if i correct 4=c. 6=d. 8=a. 7=b. and the last question would be c. because they are parallel lines and they are congruent.

step-by-step explanation:

the reason everything has to equal to 180 is because that is how it works. i'm guessing that is its rule. 4 and 8 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 6 and 7 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 60+60 is 180. so, now you have your answer

step-by-step explanation:

this can be solved first with pythagorean theorem, which tells us to square both legs of the triangle and sum it. this will be equal to the square of the hypotenuse.

the hypotenuse is the side "opposite" the 90 degree angle, which is x.

so we can write using pythagorean theorem,

simplifying:

to further simplify this radical, we need to use the property  

thus

for this case we have that by definition, the surface area of a cylinder is given by:

where:

r: it is the radius of the cylinder

h: it is the height of the cylinder

we can also rewrite it as:

where:

la: it's the lateral area

ba: area of the base

option a