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1 year ago

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Which graph represents the given equation? A. A parabola declines through (negative 1, 4), (negative 1, 3), (0, 2) and (1 point

5, negative 4 point 5) and rises through (2, negative 4), (3, 0) and (4, 6) on the x y coordinate plane. B. A parabola declines through (negative 4, 5), (negative 3 point 5, 2), (negative 3, 0) and (negative 1 point 5, negative 4 point 5) and rises through (0, negative 2), (1, 3) and (1, 4) on the x y coordinate plane. C. A parabola declines through (negative 2 point 3, 5), (negative 2, 2) and (negative 0 point 5, negative 3 point 1) and rises through (0, negative 2), (1, 4) and (1, 5) on the x y coordinate plane. D. A parabola declines through (negative 1, 5), (negative 0 point 5, 2), (0, negative 2) and (0 point 5, negative 3 point 1) and increases through (2, 2), (2 point 5, 5) on the x y coordinate plane.

1 answer:

Advocard [28]1 year ago

5 0

The graph of the quadratic function y = 1.5x² + 4x - 2 is given by the image shown at the end of the answer.

<h3>How to obtain the graph of a quadratic function?</h3>

To obtain the graph of a quadratic function, these three features have to be obtained from the function's definition:

The x-intercepts, which are the roots of the function.

The y-intercept, which is the value of y when the function crosses the y-axis.

The vertex, which is the turning point of the function.

The function for this problem is defined as follows:

y = 1.5x² + 4x - 2.

Hence the numeric values of the coefficients are listed as follows:

a = 1.5, b = 4, c = -2.

Inserting these coefficients into a calculator, the x-intercepts of the function are given as follows:

x = -3.09 -> hence the function passes through point (-3.09, 0).

x = 0.43 -> hence the function passes through point (0.43, 0).

The y-intercept of the function is given by coefficient c = -2, hence the function also passes through point (0, -2).

The x-coordinate of the vertex is given as follows:

x = -b/2a = -4/3 = -1.33.

Hence the y-coordinate of the vertex is of:

y = 1.5(-1.33)² + 4(-1.33) - 2 = -4.67.

<h3>Missing Information</h3>

The problem asks for the graph of the following function:

y = 1.5x² + 4x - 2.

More can be learned about quadratic functions at brainly.com/question/24737967

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3 years ago

The volume of a sphere is 4,000π m3. What is the surface area of the sphere to the nearest square meter? 181 m2 50,265 m2 2,614

sashaice [31]

Volume of a sphere = 4/3 πr³
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Now, surface area = 4πr² = 4 * 3.14 * (10.19)²
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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.

(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

Determine the equation of the circle with center (0,6) containing the point (74,2)

Lubov Fominskaja [6]

The circle with center (0,6) containing the point (74,2) has an equation of x² + (y - 6)² = 74.1²

<h3>How to calculate the equation of a circle</h3>

The equation of a circle is:

(x - h)² + (y - k)² = r²

Where (h, k) is the center and r is the radius.

The circle with center (0,6) containing the point (74,2). Hence:

$Radius=\sqrt{(2-6)^2+(74 -0)^2}=74.1\ unit$

The equation is:

x² + (y - 6)² = 74.1²

Find out more on circle at: brainly.com/question/24375372

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2 years ago