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Jet001 [13]
3 years ago
15

Idk what end behavior for this?

Mathematics
2 answers:
Ad libitum [116K]3 years ago
4 0

Answer:

b

Step-by-step explanation:

The end behaviour is what happens when x gets larger and positive ( right hand end ) or larger and negative ( left hand end ) Tis is called the end behaviour as x → + ∞ and x → - ∞ respectively

For a polynomial the end behaviour is determined by the term of greatest degree.

For the given function

f(x) = x^{5} - 3x³ + 2x + 4 ← degree 5 polynomial

The leading coefficient is positive

• Odd degree, positive leading coefficient, then

as x → - ∞, f(x) → - ∞

as x → + ∞, f(x) → + ∞

----------------------------------------------------------------------

• Odd degree, negative leading coefficient, then

as x → - ∞, f(x) → f(x) → + ∞

as x → + ∞, f(x) → - ∞

dangina [55]3 years ago
3 0

Answer:

It is b.

Step-by-step explanation:

When x is negative x^5 will also be negative.

f(x) = x^5 - 3x^3 + 2x + 4

As x  --> -∞ x^5 will  be the main factor for f(x) --->  -∞ .

Similarly  x^5 will have the greatest influence when x ---> ∞, so f(x) ---> ∞.

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Answer:

  5.2 in

Step-by-step explanation:

An equiangular triangle is also equilateral. All sides will be 6 inches, and the height line will be a perpendicular bisector of the side.

Using trig, the height is ...

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Using the Pythagorean theorem, ...

  h² = 6² -3² = 27

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Maksim231197 [3]

Answer: √58 cm or 7.62 cm

Step-by-step explanation:

The Pythagorean Theorem is a²+b²=c². Since we are given the lengths of a and b, we can plug them into this formula to find the length of the hypotenuse.

7²+3²=c²

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6 0
3 years ago
How would I simplify the equation 5u^6x^2+20wx/15x^4
Akimi4 [234]

Answer:

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Step-by-step explanation:

To simplify the expression given to use, we need to reduce the given expression in its simplest form. There should not be any possibility for the further cancelling out of the division terms, if any.

Now the expression that is given to us is:

5u^6x^2+\frac{20wx}{15x^4}\\

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Step-by-step explanation:

ur answer I have the link

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