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Lerok [7]
3 years ago
8

Quadrilateral ABCD is inscribed in this circle. What is the measure of angle B?

Mathematics
1 answer:
adelina 88 [10]3 years ago
6 0
Angle B measures to 140 degrees.
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Piravena must make a trip from A to B then from B to C, then from C to A. Each of these three parts of the trip is made entirely
anyanavicka [17]

Answer:

a) Cost of flying from A to B = $425

b) Total distance travelled by Piravena during the complete trip = 7,500 km

c) To minimize cost and arrive at the final cost given, she must have travelled by bus from B to C and then travelled by taking an airplane from C to A.

Step-by-step explanation:

The complete question is presented in the attached image to this question.

Full Question

a) To begin her trip she flew from A to B. Determine the cost of flying from A to B.

b) Determine the distance she travels for her complete trip.

c) Piravena chose the least expensive way to travel between cities and her total cost was $1012.50. Given that she flew from A to B, determine her method of transportation from B to C and her method of transportation from C to A.

Solution

a) The distance from A to B is given as 3250 km.

To take an airplane, it costs her a $100 booking fee, plus $0.10 per kilometer.

So, to fly 3250 km, she will pay

100 + (0.10×3250) = $425

b) For her complete journey, she is to make a trip from A to B then from B to C, then from C to A.

Her complete distance travelled = AB + BC + CA

But it is given that the three cities form a right angled triangle as given in the question with AB serving as the hypotenuse side.

Pythagoras theorem gives that the square of the hypotenuse side is equal to the sum of the respective squares of the other two sides

AB² = BC² + CA²

3250² = BC² + 3000²

BC² = 3250² - 3000² = 1,562,500

BC = √1,562,500 = 1,250 km

Total distance covered by Piravena during the entire trip = AB + BC + CA = 3250 + 1250 + 3000 = 7,500 km

c) Her total cost of travel = $1012.50

But she definitely flew from A to B at a cost of $425

This means she spent (1012.50 - 425) on the rest of the journey, that is, $587.5

Note that to travel by bus, it is $0.15 per kilometre and to travel by airplane is $100 + $0.10 per kilometre. Indicating that the airplane saves cost on long distance travels while the bus saves cost on short distance travels.

To confirm this, we calculate the two options (bus or airplane) for each route.

If she travels B to C by bus, cost = 0.15 × 1250 = $187.5

If she travels B to C by airplane, cost = 100 + (0.10×1250) = $225

Hence, the bus obviously minimizes cost here.

If she travels from C to A by bus, cost = 0.15 × 3000 = $450

If she travels from C to A by airplane, cost = 100 + (0.10×3000) = $400

Here, travelling by airplane minimizes the cost.

So, if we confirm now that she travelled from B to C by bus and then from C to A by airplane, total cost = 187.5 + 400 = $587.5

which is the remaining part of her total cost is she minimized expenses!

Hope this Helps!!!

7 0
3 years ago
Solve the system of the linear equation using any method:<br> 2x+5y=4<br> 4x+7y=11
mr_godi [17]
Multiply the first equation by 2:
4x+10y=8
Subtract the second equation from this so the 4x cancels:
3y = -3
y = -1
Plug into first equation:
2x -5 = 4
2x = 9
x = 9/2 = 4 1/2 = 4.5
So (x,y) = (-1,4.5)
3 0
3 years ago
Read 2 more answers
A project is graded on a scale of 1 to 5. If the random variable, X, is the project grade, what is the mean of the probability
alex41 [277]

Answer:

the mean is 3

answer choice D

Step-by-step explanation:

60/20

8 0
3 years ago
Read 2 more answers
Little Melinda has nickels and quarters in her bank. She has two
SCORPION-xisa [38]

A) N +2 = Q which equals

A) N -Q = -2

B) .05N +.25Q = 3.50  multiplying A) by .25

A) .25N -.25Q = -.5  then adding A) and B)

.30N = 3

Nickels = 10 Quarters = 12

*************DOUBLE CHECK ***************

.05 nickels = $0.50  .25 Quarters = $3.00





7 0
3 years ago
Can i have help with these questions
adell [148]

Answer:

  (-3, 5), (-1, -1), (5, -3)

Step-by-step explanation:

Each pair of vertices can be one of the diagonals. Then the missing point will be found at the coordinates that are the sum of those, less the coordinates of the third point.

Given points are ...

  A(-2, 2), B(1, 1), C(2, -2)

For AB a diagonal, D1 is ...

  A+B-C = (-2+1-2, 2+1-(-2)) = (-3, 5)

For AC a diagonal, D2 is ...

  A+C-B = (-2+2-1, 2-2-1) = (-1, -1)

For BC a diagonal, D3 is ...

  B+C-A = (1+2-(-2), 1-2-2) = (5, -3)

_____

For a lot of parallelogram problems I find it easiest to work with the fact that the diagonals bisect each other. This means they both have the same midpoint, so for quadrilateral ABCD, we have (A+C)/2 = (B+D)/2. Multiplying this by 2 gives the equation we used above, A+C = B+D, so D=A+C-B. Remember, in ABCD, AC and BD are the diagonals.

6 0
2 years ago
Read 2 more answers
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