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4 years ago

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23. Critical Thinking Lucia knows the fourth term in a sequence is 55 and the ninth term in the same sequence is 90. Explain how

she can find the common difference for the sequence. Then use the common difference to find the second term of the sequence.

1 answer:

Anarel [89]4 years ago

4 0

Let's call <em>a</em> the first term and <em>d</em> the common difference.

$a_n = a+(n-1)d$

The fourth term is 55:

$a_4 = 55 = a + 3d$

The ninth term is 90

$a_9 = 90 = a + 8d$

Subtracting,

$90 - 55 = 8d - 3d$

$35 = 5d$

$d = 7$

Answer: common difference of 7

We found the common difference. Substituting,

$a = 55 - 3d = 55 - 21 = 34$

Second term

$a_2 = a + d = 34 + 7 = 41$

Answer: second term of 41

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During the basketball season, Cory made 21 of the 60 baskets she attempted. Krista made 16 out of 40 baskets she attempted. Paul

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Given:

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Solution:

We know that,

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3 years ago

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

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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.

(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:

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

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.

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4 years ago