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zlopas [31]
3 years ago
7

What theorems can be utilized in solving polynomials and their depressed equations? How can polynomial functions be written when

given the zeros?
Mathematics
1 answer:
ki77a [65]3 years ago
7 0

Answer:

Fundamental theorem of algebra, rational root theorem, Descartes rule of signs, remainder theorem, and the factor theorem can be utilized in solving polynomials and their depressed equations.  

Polynomial functions can be written when given the zeros by determining the factors of the function, and multiplying the factors.

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Find the equation of the line that passes through (4,4) and is parallel to
Black_prince [1.1K]

Answer:

y = - 3x + 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

3x + y + 2 = 0 ( subtract 3x + 2 from both sides )

y = - 3x - 2 ← in slope- intercept form

with slope m = - 3

Parallel lines have equal slopes, thus

y = - 3x + c ← is the partial equation

To find c substitute (3, - 4) into the partial equation

- 4 = - 9 + c ⇒ c = - 4 + 9 = 5

y = - 3x + 5 ← equation of line in form y = mx + c

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3 years ago
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Answer:

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8 0
3 years ago
Using a Celsius scale, water freezes at 0°C and boils at 100°C. Using a Fahrenheit scale, water freezes at 32°F and boils at 212
natulia [17]
Y= (x(9/5)) + 32; Ex: y=(100(9/5)) + 32 —> 100 x 1.8 (or 9/5) = 180–> 180 + 32= 212 F
5 0
3 years ago
Can someone help me im stuck
Tcecarenko [31]

Answer:

<em>(3). 500; (4) 60</em>

Step-by-step explanation:

5 0
3 years ago
Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun
WITCHER [35]

Answer:

Required solution (a) f(x,y,z)=xyz+3z^2+C (b) 40.

Step-by-step explanation:

Given,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k

(a) Let,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}

Then,

f_x=yz,f_y=xz,f_z=xy+6z

Integrating f_x we get,

f(x,y,z)=xyz+g(y,z)

Differentiate this with respect to y we get,

f_y=xz+g'(y,z)

compairinfg with f_y=xz of the given function we get,

g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)

Then,

f(x,y,z)=xyz+h(z)

Again differentiate with respect to z we get,

f_z=xy+h'(z)=xy+6z

on compairing we get,

h'(z)=6z\implies h(z)=3z^2+C   (By integrating h'(z))  where C is integration constant. Hence,

f(x,y,z)=xyz+3z^2+C

(b) Next, to find the itegration,

\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40

3 0
3 years ago
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