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2 years ago

The graph of F(x), shown below, resembles the graph of G(x) = x2, but it has

Mathematics

1 answer:

DENIUS [597]2 years ago

6 0

F(x)= 2(x-2)^2 + 2 is the correct equation of F(x)

A quadratic function is a polynomial function of the second degree. The general form of a quadratic function is this: f (x) = ax^2 + bx + c, where a, b, and c are real numbers and a≠ 0.

Graphs of quadratic functions

The term "parabola" refers to the graph of a quadratic function. A parabola essentially resembles the letter "U," yet it can be inverted or exactly this shape depending on the situation. The leading coefficient determines whether the graph of a quadratic function opens up or down; if it is more than zero, the parabola opens up; if it is less than zero, the parabola opens down.

It is given that graph of F(x), shown below, resembles the graph of G(x) = x^2, but it has been changed somewhat.

We need to find the equation of F(x)

In the figure, parabola opens up, the leading coefficient is greater than zero ,So option (d) is correct

Hence the equation of F(x) is 2(x-2)^2 + 2

Learn more about parabola here:

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In a mixture of concrete, there are five lb of cement mix for each pound of gravel. if the mixture contains a total of 198 lb o

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There are 39.6 lbs. of gravel.

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3 years ago

Find the value of a and b if root3-1/root3+1+ root3+1/root3-1=a+root3 b

Digiron [165]

Answer:

a=4 and b = 0

Step-by-step explanation:

Given : $\dfrac{\sqrt 3 -1}{\sqrt 3+1}+\dfrac{\sqrt 3+1}{\sqrt 3-1}=a+\sqrt 3 b$

To find, The value of a and b

Solution,

Solving LHS of the given equation,

$\dfrac{(\sqrt 3-1)^2+(\sqrt3+1)^2}{(\sqrt 3+1)(\sqrt 3-1)}=a+\sqrt 3 b$

Since,

$(a-b)^2=a^2+b^2-2ab\\\\(a+b)^2=a^2+b^2+2ab\\\\(a-b)(a+b)=a^2+b^2$

So,

$\dfrac{3+1-2\sqrt 3+3+1+2\sqrt 3}{3-1}=a+\sqrt 3 b\\\\4=a+\sqrt 3 b$

$4+0=a+\sqrt 3 b$

On comparing we get :

a = 4 and b = 0

4 0

3 years ago

I need quick help pls!! by reason: AAA or ASA, SAA or Cannot be determined or SAS or SSS

Ray Of Light [21]

Answer:

ΔDCE by ASA

Step-by-step explanation:

The marks on the diagram show AE ≅ DE. We know vertical angles AEB and DEC are congruent, and we know alternate interior angles BAE and CDE are congruent. The congruent angles we have identified are on either end of the congruent segment, so the ASA theorem applies.

Matching corresponding vertices, we can declare ΔABE ≅ ΔDCE.

5 0

4 years ago

45 is 20% of what number?

nadezda [96]

$x-\ number\\\\
20\%x=45\\\\
\frac{20}{100}x=45\\\\
0.2x=45\ \ \ | divide\ by\ 0.2\\\\
x=225\\\\
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5 0

3 years ago

find the scale, k, of H(3,4) if H'(9,12) after going through the transformation (x,y) ---> (kx,ky) centered at the origin

stiks02 [169]

The value of scale k is 3

Step-by-step explanation:

In order to find the scale , we can either divide the coordinates of H' by the original point or compare both the points to find the original ordered pair.

Given

H(3,4)

H'(9,12)

The coordinates of H are multiplied with some integer k to form H'

So,

For x-coordinate:

3k = 9

k = 9/3 = 3

For y-coordinate:

4k = 12

k = 12/4 = 3

Hence,

The value of scale k is 3

Keywords: Coordinate geometry, scaling

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5 0

3 years ago