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3 years ago

Create a scale model of the solar system to show relative sizes os the sun and the planets

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

7 0

Answer:

Ross

Step-by-step explanation:

hi ki nhi h Uske utna hi nhi h Uske to be a good time to meet you at your earliest reply to

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For each, list three elements and then show it is a vector space.

Butoxors [25]

Answer:

(a) Three polynomials of degree 1 with real coefficients belong to the set $P_1=\{a_0+a_1x\ | a_0, a_1 \in \mathbb{R} \}$ , then:

$2+3x \in P_1$

$4.5+\sqrt2 x \in P_1$

$\log5+78x \in P_1$

(b) Three polynomials of degree 1 with real coefficients that hold the relation $a_0 - 2a_1 = 0$ belong to the set $P_2=\{a_0+a_1x\ | a_0-2 a_1 =0 \}$ . The relation between the coefficients is equivalent to $a_1 = \frac{a_0}{2}$ , then:

$4+2x \in P_2$

$13+6.5x \in P_2$

$10.5+5.25x \in P_2$

Step-by-step explanation:

(a) Three polynomials of degree 1 with real coefficients belong to the set $P_1=\{a_0+a_1x\ | a_0, a_1 \in \mathbb{R} \}$ , then:

$2+3x \in P_1$
$4.5+\sqrt2 x \in P_1$
$\log5+78x \in P_1$

A vector space is any set whose elements hold the following axioms for any $\vec{u}, \vec{v}$ and $\vec{w}$ and for any scalar $a$ and $b$ :

$(\vec{u} + \vec{v} )+\vec{w} = \vec{u} +( \vec{v} +\vec{w})$
There is the <em>zero element </em>such that: $\vec{0} + \vec{u} = \vec{u} + \vec{0}$
For all element $\vec{u}$ of the set, there is an element $-\vec{u}$ such that: $-\vec{u} + \vec{u} = \vec{u} + (-\vec{u}) = \vec{0}$
$\vec{u} + \vec{v} = \vec{v} + \vec{u}$
$a(b\vec{v}) = (ab)\vec{v}$
$1\vec{u} = \vec{u}$
$a(\vec{u} + \vec{v} ) = a\vec{u} + a\vec{v}$
$(a+b)\vec{v} = a\vec{v}+b\vec{v}$

Let's proof each of them for the first set. For the proof, I will define the polynomials $a_0+a_1x$ , $b_0+b_1x$ and $c_0+c_1x$ and the scalar $h$ and $g$ .

$(a_0+a_1x + b_0+b_1x)+c_0+c_1x = a_0+a_1x +( b_0+b_1x+c_0+c_1x)\\(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x$ and defining $a_0+b_0+c_0 = \alpha_0$ and $a_1+b_1+c_1 = \alpha_1$ , we obtain $\boxed{\alpha_0+\alpha_1x= \alpha_0+\alpha_1x}$ which is another polynomial that belongs to $P_1$
A null polynomial is define as the one with all it coefficient being 0, therefore: $\boxed{0 + a_0+a_1x = a_0+a_1x + 0 = a_0+a_1x}$
Defining the inverse element in the addition as $-a_0-a_1x$ , then $-a_0-a_1x + a_0 + a_1x = a_0+a_1x + (-a_0-a_1x)\\\boxed{(-a_0+a_0)+(-a_1+a_1)x = (a_0-a_0)+(a_1-a_1)x = 0}$
$(a_0+a_1x) +( b_0+b_1x) =( b_0+b_1x) +( a_0+a_1x)\\(a_0+b_0)+(a_1+b_1)x = (b_0+a_0)+(b_1+a_1)x\\\boxed{(a_0+b_0)+(a_1+b_1)x = (a_0+b_0)+(a_1+b_1)x}$
$a[b(a_0+a_1x)] = ab (a_0+a_1x)\\a[ba_0+ba_1x] = aba_0+aba_1x\\\boxed{aba_0+aba_1x = aba_0+aba_1x}$
$\boxed{1 \cdot (a_0+a_1x) = a_0+a_1x}$
$\boxed{a[(a_0+a_1x)+(b_0+b_1x)] = a(a_0+a_1x) + a(b_0+b_1x)}$
$(a+b)(a_0+a_1x)=aa_0+aa_1x+ba_0+ab_1x\\\boxed{(a+b)(a_0+a_1x)= a(a_0+a_1x) + b (a_0+a_1x)}$

With this, we proof the set $P_1$ is a vector space with the usual polynomial addition and scalar multiplication operations.

$4+2x \in P_2$
$13+6.5x \in P_2$
$10.5+5.25x \in P_2$

Let's proof each of axioms for this set. For the proof, I will define again the polynomials $a_0+a_1x$ , $b_0+b_1x$ and $c_0+c_1x$ and the scalar $h$ and $g$ . Again the relation $a_1 = \frac{a_0}{2}$ between the coefficients holds

$[(a_0+a_1x) +( b_0+b_1x)]+(c_0+c_1x) = (a_0+a_1x) +[( b_0+b_1x)+(c_0+c_1x)]\\(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x$ and considering the coefficient relation and defining $a_0+b_0+c_0 = \alpha_0$ and $a_1+b_1+c_1 = \alpha_1$ , we have $(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x\\(a_0+b_0+c_0) + \frac{1}{2} (a_0+b_0+c_0)x = (a_0+b_0+c_0) + \frac{1}{2} (a_0+b_0+c_0)x\\\boxed{\alpha_0 + \alpha1x = \alpha_0 + \alpha1x}$ which is another element of the set since it is a degree one polynomial whose coefficient follow the given relation.

The proof of the other axioms can be done using the same logic as in (a) and checking that the relation between the coefficients is always the same.

6 0

4 years ago

Please answer this sample space question.

Tju [1.3M]

I think the answer is c 24

7 0

3 years ago

-3а - b - 3c = 9 <br>4а + 4 - 4c = 4 <br>2а + 4c = 18

Reil [10]

Answer:

0is the answer to this question

4 0

3 years ago

Given the expression -15 (10 2/3) A. Rewrite the expression using distributive property B. Evaluate the expression in show all y

Ainat [17]

9514 1404 393

Answer:

-160

Step-by-step explanation:

-15(10 2/3) = -15(10) +(-15)(2/3) = -150 -30/3 = -150 -10 = -160

The mixed number can be written as the sum of an integer and a fraction. The multiplier then multiplies the integer and the fraction separately, and the two products are added.

8 0

3 years ago

Could someone pls help me !!! do t make fun of me just cuz it’s easy for u!!

pantera1 [17]

9+t=56

You are supposed to minus 9

9-9+ t= 56-9

t= 47

check answer:

9+47=56

56=56

t= 47 is the right answer

4 0

3 years ago

Read 2 more answers