All categories

3 years ago

According to the fundamental theorem of algebra, how many zeros does the polynomial below have?

Mathematics

2 answers:

madam [21]3 years ago

7 0

Answer: 3!

Step-by-step explanation:

A P E X

Oxana [17]3 years ago

6 0

It has 3 zeros, because its a polynomial of degree 3.

You might be interested in

What is the slope–intercept form of the equation 4x − 2y = 9?

adelina 88 [10]

Hey there! :D

Slope intercept form: y=mx+b

4x-2y=9

Add 2y to the other side.

4x=9+2y

Subtract 9 on both sides.

4x-9=2y

Flip the equation over.

2y=4x-9 <== slope intercept form

I hope this helps!
~kaikers

5 0

3 years ago

9 and 5 twelves, minus 6 and 7 twelves

mash [69]

Answer:

2 5/6

Step-by-step explanation:

9 5/12 - 6 7/12

We will need to borrow from the 9

8 12/12 + 5/12 - 6 7/12

8 17/12 - 6 7/12

(8-6) + (17/12 -7/12)

2 10/12

This simplifies. Divide the top and bottom of the fraction by 2

2 5/6

5 0

3 years ago

Help thanks 7.04 love u guys

Ugo [173]

Answer:

D. $-\frac{3}{4}$

Step-by-step explanation:

The given equation is

$4y=8x-3$

When we divide through by 4 we get;

$y=2x-\frac{3}{4}$

Comparing to

y=mx+c,

The y-intercept is $-\frac{3}{4}$

The corrrect choice is D.

3 0

3 years ago

Read 2 more answers

How would you simplify the expression represented by the diagram below? Four figures. The first consists of 4 squares each label

Sergeu [11.5K]

Answer:

6 x 3c

Step-by-step explanation:

6 0

3 years ago

Kamal wrote the augmented matrix below to represent a system of equations

ra1l [238]

Consider the operation is $R_2\to -3R_2$ .

Given:

The augmented matrix below represents a system of equations.

$\left[\left.\begin{matrix}1&0&1\\1&3&-1\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-9\\-2\end{matrix}\right]$

To find:

Matrix results from the operation $R_2\to -3R_2$ .

Step-by-step explanation:

We have,

$\left[\left.\begin{matrix}1&0&1\\1&3&-1\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-9\\-2\end{matrix}\right]$

After applying $R_2\to -3R_2$ , we get

$\left[\left.\begin{matrix}1&0&1\\-3(1)&-3(3)&-3(-1)\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-3(-9)\\-2\end{matrix}\right]$

$\left[\left.\begin{matrix}1&0&1\\-3&-9&3\\3&2&0\end{matrix}\right|\begin{matrix}-1\\27\\-2\end{matrix}\right]$

Therefore, the correct option is A.

3 0

3 years ago