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3 years ago

What is the vertex of the graph of f(x) = |x-3| + 11?

Mathematics

2 answers:

Serga [27]3 years ago

6 0

Answer:

(3,11)

Step-by-step explanation:

we find the vertex of

$g(x)=|x-3|$ which is going to be (x,0)

and we add 11 (movement upward) so the vertex is (x,11)

vertex of g(x)

it is going to be when g(x)=0

and that only happens when x=3

so vertex (3,0)

so vertex of f(x)

(3,11)

if you don't believe i leave you the graph

weqwewe [10]3 years ago

4 0

The function f(x) = |x-3| +11 is in the vertex form which means vertex is (3,11)

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Olegator [25]

Answer:

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Step-by-step explanation:

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3 years ago

addison and kelsey are running on a path modeled by x^2+y^2-10x-18y-378=0, where the distance is in meters. what is the maximum

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The equation of a circle of radius r and center $(x_0,y_0)$ is given by:

$(x - x_0)^2 + (y - y_0)^2 = r^2$

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In this problem, the circular path is modeled by:

$x^2 + y^2 - 10x - 18y - 378 = 0$

We complete the squares to place it in the standard format, thus:

$x^2 - 10x + y^2 - 18y = 378$

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Thus, the radius is:

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The maximum distance is the diameter of the circle, which is of 44 units.

A similar problem is given at brainly.com/question/24992361

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2 years ago

30 POINTS! Combine the like terms of the 2 questions below:

astraxan [27]

Answer:

$\tt 1)\: \boxed{\tt 4y+10}$

$\tt 2)\:\boxed{\tt 0.5x^4+10.5}$

Step-by-step explanation:

1) y + 4 + 3(y + 2)

Expand:-

$\tt y+4+3y+6$

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$\tt 0.5x^4-1.5+12$

Add/Subtract Numbers:-

$\boxed{\tt 0.5x^4+10.5}$

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2 years ago

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Kay [80]

Answer:

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