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djyliett [7]
2 years ago
11

Need help on this please

Mathematics
1 answer:
Effectus [21]2 years ago
6 0
Area of a circle is 3.14r^2
radius is 7 (half of 14)
3.14(7)^2
=153.9
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Which of the following functions is graphed below
Ratling [72]

The function that is graphed is y = |x -2| + 3. The correct option is D. y = |x -2| + 3

<h3>Absolute value graph</h3>

From the question, we are to determine the function for the given graph

From the graph,

When x = 2, y = 3

The equation that satisfy this condition is

y = |x -2| + 3

NOTE: This function satisfies all the points on the graph

Hence, the function that is graphed is y = |x -2| + 3

Learn more on Absolute value graph here: brainly.com/question/17082317

#SPJ1

8 0
1 year ago
Find the radius if the endpoints of the diameter are (-6,5) and (2, 3).
Fiesta28 [93]

Answer:

x

−

a

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2

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y

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Step-by-step explanation:

6 0
2 years ago
Janet was in an accident. The damage to her car is
Svet_ta [14]

hi tAnswer:hey

Step-by-step explanation:

7 0
2 years ago
(3w-8)(5=w)=0 ,What is w?
Lemur [1.5K]

Answer:

- 1. 5 as decimal

-3 / 2 as a fraction

Step-by-step explanation:

7 0
3 years ago
A triangle has vertices at (1,10), (-5,2), and (7,2). What is its orthocenter. Show your work.
kondor19780726 [428]

Answer:

Question: What is the orthocenter of a triangle with the vertices (-1,2) (5,2) and (2,1)?

The coordinates of point A are (-1,2), point B are (5,2), and point C are (2,1).

The orthocent is the intersection of the three altitudes. An altitude goes from a vertex and is perpendicular to the line containing the opposite side.

In the coordinate plane the equations of the altitudes can be found and then a system of equations can be solved.

Altitude 1. From point C perpendicular to the line containing side AB.

Slope of line AB is 0 (horizontal line), a vertical line is perpendicular to a horizontal line. Thus, the equation of altitude 1 is  x=2 .

Altitude 2. From point B perpendicular to the line containing side AC.

Slope of line AC is  −13 , the slope of a line perpendicular to line AC is 3. The equation of altitude 2 is  y=3x−13  

Altitude 3. From point A perpendicular to the line containing side BC.

Slope of line BC is  13 , the slope of a line perpendicular to line BC is  −3 . The equation of altitude 3 is  y=−3x−1  

The orthocenter is the point where all three altitudes intersect.

x=2  

y=3x−13  

y=−3x−1  

Use substitution to solve the first two equations  y=3(2)−13=−7  

The orthocenter is the point  (2,−7)  

we did not need the third equation, but we can use it as a check, plug the coordinates into the third equation:

−7=−3(2)−1  

−7=−6−1  

−7=−7  it works.

3 0
2 years ago
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