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3 years ago

Can this expression be simplified? If not, explain why. WILL MARK BRAINLIEST.

Mathematics

2 answers:

irakobra [83]3 years ago

8 0

Answer:

yes

Step-by-step explanation:

see attached graph - there are two real roots for y=2x^4-6x^2+10x-7

so 2x^4-6x^2+10x-7 can be expressed as:

(x - root1)(x - root2)(2x^2+Bx+C)

Arada [10]3 years ago

4 0

Answer:

Yes, it can be.

Step-by-step explanation:

Using solver program, the expression 2x^4-6x^2+10x-7 has:

two real roots at x≈-2.3888 and x≈1.1424

and

two complex roots at x≈0.62321 - 0.94559 i and x≈0.62321 + 0.94559 i

Because it has two real roots, the expression can be simplified.

You might be interested in

Anyone help?<br><br> What is the derivative of x^2 + 2x - 4?<br><br> Thanks!

castortr0y [4]

Let's take the derivative of each term one by 1

x^2
Apply the power rule for derivatives.
2x^(2-1) = 2x
You get 2x.

2x
Apply the power rule (x^1) and the constant multiple rule.
2 * x^(1-1) = 2 * x^0 = 2
You get 2.

-4
The derivative of a constant is 0.

Put that together and you get 2x + 2

Have an awesome day! :)

4 0

3 years ago

Read 2 more answers

21 x ________ ÷ 3 = 21 *

GrogVix [38]

Answer:

first one: <u>3</u>

second one: <u>b</u>

third one: <u>3.2</u>

Step-by-step explanation:

first one:

21 x ________ ÷ 3 = 21

rewrite equation:

21 ÷ 3 x ___ = 21

7 x ___ = 21

(21 / 7 = 3)

7 x 3 = 21

second one:

b x h ÷ h = ___

third one:

876 x 3.2 ÷ ______ = 876

2803.2 ÷ ___ = 876

(2803.2 / 876 = 3.2)

2803.2 x 3.2 = 876

7 0

3 years ago

the sum of three numbers is 144. the third number is 4 times the first. the second number is 6 more than the first. what are the

Law Incorporation [45]

First I would change the descriptions of the numbers into expressions.

first number is n
second number is n + 6
third number is 4n (4 x n)

Then you would insert these expressions into an equation and isolate n.

n + n + 6 + 4n = 144
n + n + 4n = 144 - 6
6n = 138
n = 138/6
n = 23

Lastly, you would plug in this value into all of the expressions.

first number is 23
second number is 23 + 6 = 29
third number is 4(23) = 92

Therefore, the numbers are 23, 29, and 92.

7 0

3 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.