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3 years ago

For brainiest:):):)):)

Mathematics

2 answers:

Reika [66]3 years ago

6 0

Answer:

Step-by-step explanation:

gtnhenbr [62]3 years ago

5 0

Answer:

1)50.24 2) 254.34

Step-by-step explanation:

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What is the quotient? 1052 divided by 52

Liono4ka [1.6K]

To find the quotient, we can use a calculator. Problem is that we get a decimal, so 52 won't divide evenly. When using a calculator the whole number part is the quotient.

1052 + 52 = 20.2307692....

Now, 52 x 20 = 1040. Since 1052 - 1040 = 12 we can properly report the quotient and its remainder.

So 1052 + 52 = 20 (its quotient) and 12 (its remainder).

7 0

3 years ago

If f(x)=(3×+7)^2, then f(1)=?

ELEN [110]

F(x) = (3x+7)^2

f(1) = (3*1+7)^2 ... replace every x with 1

f(1) = (3+7)^2

f(1) = 10^2

f(1) = 100

Final Answer: 100

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3 years ago

Read 2 more answers

Tim has 74 socks, all in pairs. He has 1 more pair of white socks than black socks. How many individual socks of each color does

tatuchka [14]

Answer:

36 black socks and 38 white socks

Step-by-step explanation:

7 0

3 years ago

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.