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3 years ago

What is the vertex for the following quadratic? f(x)=2(x - 4)^2 -8

Mathematics

1 answer:

Ghella [55]3 years ago

6 0

Answer:

vertex = (4, - 8 )

Step-by-step explanation:

The equation of a quadratic in vertex form is

f(x) = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

f(x) = (x - 4)² - 8 ← is in vertex form

with (h, k ) = (4, - 8 ) ← coordinates of vertex

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Evaluate the expression.<br><br> (13^2 - 12^2) ÷ 5

otez555 [7]

Answer:

5

Step-by-step explanation:

(13^2 - 12^2) ÷ 5

(169 - 144) ÷ 5

25 ÷ 5

5

5 0

2 years ago

Note that WXYZ has vertices W(-1, 2), X(-5, 7), y(-1, -2), and Z (3, -7).

svp [43]

a. Slope of WZ = -2.25; Slope of WX = 5

b. WZ = √97; WX = √41

c. WXYZ is not a <em>rectangle, rhombus, nor a square</em>. We can conclude that: <em>D. WXYZ is none of these</em>.

<h3>Slope of a Segment</h3>

Slope = change in y/change in x

Given:

W(-1, 2), X(-5, 7), Y(-1, -2), and Z (3, -7)

a. Slope of WZ and slope of WX:

Slope of WZ = (-7 - 2)/(3 -(-1)) = -2.25

Slope of WX = (7 - 2)/(-1 -(-1)) = 5

b. Use distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , to find WZ and WX:

$WZ = \sqrt{(3 - (-1))^2 + (-7 - 2)^2}\\\\\mathbf{WZ = \sqrt{97} }$

$WX = \sqrt{(-5 -(-1))^2 + (7 - 2)^2}\\\\\mathbf{WX = \sqrt{41} }$

c. The quadrilateral WXYZ have adjacent sides that are not perpendicular to each other and have different slopes and different lengths, so therefore, WXYZ is not a rectangle, rhombus, nor a square. We can conclude that: <em>D. WXYZ is none of these</em>.

Learn more about slopes on:

brainly.com/question/3493733

5 0

2 years ago

Write a quadratic equation with the given roots. Write the equation in the form of ax^2+bx+c=0 where a b and c are integers

Bas_tet [7]

You haven't provided the required roots, but I can tell you how to do this kind of exercises in general.

If the $x^2$ coefficient is 1, i.e. the equation is written like $x^2+bx+c=0$ , then you can say the following about the coefficients b and c:

$b$ is the opposite of the sum of the roots
$c$ is the multiplication of the roots.

So, for example, if we want an equation whose roots are 4 and -2, we have:

$4+(-2) = 4-2 = 2 \implies b = -2$
$4 \cdot (-2) = -8 \implies c = -8$

So, the equation is $x^2-2x-8=0$

If your roots are rational, you can work like this: suppose you want an equation with roots 3/4 and 1/2. You have:

$\dfrac{3}{4}+\dfrac{1}{2} = \dfrac{3}{4}+\dfrac{2}{4} = \dfrac{5}{4} \implies b = -\dfrac{5}{4}$
$\dfrac{3}{4} \cdot \dfrac{1}{2} = \dfrac{3}{8} \implies c = \dfrac{3}{8}$