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uysha [10]
2 years ago
13

Find the unknown angle for q, r, and t​

Mathematics
1 answer:
Sav [38]2 years ago
3 0

Answer:

r=90

q=45?

t=52

Step-by-step explanation:

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a balloon dress 140 miles towards the lessons 45 seconds then it suddenly changes and the win flies 90 miles towards the east an
Bezzdna [24]
I think the total is 190 miles they have traveled but don’t trust me I think it is
3 0
3 years ago
16
Tanya [424]

Answer:

2.50

Step-by-step explanation:

Given

(x_1,y_1) = (-1,0.8)

(x_2,y_2) = (0,2)

(x_3,y_3) = (1,5)

(x_4,y_4) = (2,12.5)

Required

The rate of change (b)

The above graph is represented as:

y = ab^x

For: (x_2,y_2) = (0,2);

We have:

y = ab^x

2 = a * b^0

2 = a *1

2 = a

a = 2

For (x_3,y_3) = (1,5),

We have:

y = ab^x

5 = a * b^1

5 = a * b

Substitute a = 2

5 = 2 * b

Divide by 2

2.5 = b

b = 2.5

<em>Hence, the rate of change is 2.50</em>

7 0
3 years ago
Y = 9 - 1 <br> What is this ?
lesya692 [45]
Y=8

your teacher is introducing algebra
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3 years ago
Q 14. Two friends Tina and Reeta simplified two different expressions during the revision
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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
3 years ago
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