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Nuetrik [128]
4 years ago
8

What is the degree of the polynomial function f(x) = 2x3 + 7x5 − 4x − 12?

Mathematics
2 answers:
djverab [1.8K]4 years ago
5 0

Answer:

degree of the polynomial is 5.

Step-by-step explanation:

We have been given the polynomial f(x)=2x^3+7x^5-4x-12

We can rearrange the polynomial and write in standard form as

f(x)=7x^5+2x^3-4x-12

For the degree of the polynomial we see the highest exponent of the polynomial.

Degree = highest exponent

The highest exponent of the polynomial is 5.

Therefore, the degree of the polynomial is 5.

-Dominant- [34]4 years ago
4 0
IF f(x) = 2x3 + 7x5 − 4x − 12 << not written to standards 
<span>IS f(x) = 7x^5 +2x^3− 4x − 12 </span>
<span>x is to the 5th power </span>
<span>then 5</span>
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PSYCHO15rus [73]

Answer:

27 & 72

Step-by-step explanation:

3*24=72 and 3*9=27 so they are both divisible by 3 and also if you subtract 24 from 72 you get 45. get it now?

4 0
3 years ago
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What are the zeros of f(x)=x^2+x-20?
zzz [600]
F(x) = x² + x - 20 = x² + 5x - 4x - 20 = x(x + 5) - 4(x + 5) = (x + 5)(x - 4)

f(x) = 0 ⇔ (x + 5)(x - 4) = 0 ⇔ x + 5 = 0 or x - 4 = 0 ⇒ x = -5 or x = 4

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3 0
3 years ago
Point T(-9,5) lies on the perpendicular bisector of UV. If the
Flura [38]

The coordinate of point V is at (-16, 9)

If Point T(-9,5) lies on the perpendicular bisector of UV, this means that the point divides the line UV into two equal parts

Given the following coordinates

Midpoint T = (-9, 5)

U = (-2, 1)

Required

coordinate of point V

Using the midpoint formulas;

T(x, y) = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \\T(-9, 5) = (\frac{-2+x_2}{2}, \frac{1+y_2}{2})

Get the value if x₂ and y₂

-9 = \frac{-2+x_2}{2}\\-18 = -2+x_2\\x_2 = -18+2\\x_2 = -16\\

Similarly;

5 = \frac{1+y_2}{2}\\10 = 1+y_2\\y_2 = 10-1\\y_2 = 9\\

Hence the coordinate of point V is at (-16, 9)

Learn more here: brainly.com/question/18049211

5 0
3 years ago
Pls i need the help.<br>​
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Answer:

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Step-by-step explanation:

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3 years ago
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Leviafan [203]

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= 0.16 foot square

4 0
3 years ago
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