What are the x-intercepts of the parabola? graph of parabola falling from the left, passing through 2 comma 3 to about 4 comma negative 1, and rising to the right, passing through 6 comma 3 (0, 3) and (0, 5) (0, 4) and (0, 5) (3, 0) and (5, 0) (4, 0) and (5, 0)

answer: a

step-by-step explanation:

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.

we need to see the graph to know but

strong positive is if it is going up from left to right in a straight diagonal line (like this /)

strong negative is the same but from right to left (like this \)

weak negative is the same as strong negative but its more straight line still right to left

and weak positive is the same as strong positive but more a straight line still left to right though.

  b:   ∠x = 28.6°, ∠y = 31.2°, ∠z = 120.2°

step-by-step explanation:

this is one of those questions where "test taking skill" is all you need. the only answer that has angles that sum to 180° is the one shown above. the remaining choices do not describe the angles of a triangle.

if you want to do more work than simply adding the offered numbers, you can also check them against the law of sines. it should be true that

  x/sin(x) = y/sin(y) = z/sin(z)

and all of these ratios must be greater than 15, the longest side length.

__

if you want to solve the triangle from scratch, i'd suggest solving for the largest angle using the law of cosines.

  z² = x² +y² -2xy·cos(z)

  z = arccos((x² +y² -z²)/(2xy)) = arccos(-75.11/149.4) = 120.182°

now you can find y from the law of sines

  y = arcsin(y/z·sin(z)) = 31.242°