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fiasKO [112]
3 years ago
13

HELP ME PLEASE NOW!!!!

Mathematics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

Calculate the slope in the four intervals with the formula

m = (f(b) - f(a)) / (b - a) slope in Intervall [a; b]

m1 = (3 - 0) / (2 - 0) = 1.5

m2 = (11 - 3) / (4 - 2) = 4

m3 = (23 - 3) / (6 - 2) = 5

m4 = (23 - 11) / (6 - 4) = 6

between x = 4 and x = 6 is the correct answer.


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Solve for x<br>2x²-3x-209=0​
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Answer:

x= 11

Step-by-step explanation:

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(-1.5, 0), (0, -1)

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You purchase a car using a $20,000 loan with a 5% simple interest rate.
hammer [34]
Simple interest = Cost Price + (Interest Percentage of Cost Price × number of years or months we are paying off)

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When building a roof, carpenters put vertical boards every 2 feet along the horizontal baseline. The height of the boards, y, fo
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10 feet

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we are trying to figure out what x is since we know the height of the board

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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
3 years ago
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