Will mark brainliest!
gus wants to determine the end behavior of f(x)=−5x^(4)+3x^(2)−x. he realizes that since the leading coefficient is _[blank 1]_ and the exponent is _[blank 2]_, as x approaches infinity, f(x) approaches negative infinity, and as x approaches negative infinity, f(x) approaches negative infinity.
match each blank with the word that correctly fills it in.
blank 1
blank 2
1. negative
2. odd
3. increasing
4. even
5. decreasing
6. positive

Blank 1:negative

Blank 2:even

Negative and Even

Step-by-step explanation:

Since the exponent is even, it makes the end behaviors in both the positive and negative direction go the same way. The reason this occurs is because a number raised to an even exponent cannot be negative. Since it is always positive that means that the sign determines if the end behavior is positive or negative and the sign of the leading coefficient is constant through this equation. Since we now know that the sign will affect the direction we can look and see the sign is negative indicating that the end behaviors for x approaching positive and negative infinity is negative infinity.

(c, 0) i might be wrong so don't put full faith on my answer

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this