All categories

3 years ago

In the figure, the three segments are tangent to the circle at points B, F, and G. If y=2/3x, find x, y, and z.

Mathematics

1 answer:

Sergio [31]3 years ago

3 0

We know that

y=(2/3)x
in the tangent at the circle at the point G
x+y=45 in-------> x+(2/3)x=45-------> (5/3)x=45------> x=45*3/5----> x=27 in
y=45-x-----> y=45-27------> y=18 in

in the tangent at the circle at the point B
z+y=39
z=39-y-------> z=39-18-----> z=21 in

in the tangent at the circle at the point F
z+x=48
so
21+27=48
so
48=48------> is ok

the answer is
x=27 in
y=18 in
z=21 in

You might be interested in

The area of a square is represented by the expression 16x^4y^8. What is the length of each side of this square

Leona [35]

Answer: $4x^2y^4$

Step-by-step explanation:

Given

The area of the square is $A=16x^4y^8$

Suppose a is the side length of the square

The area of the square is given by

$\Rightarrow A=(\text{side length})^2$

So, we can write

$\Rightarrow a^2=16x^4y^8\\\Rightarrow a=4x^2y^4$

length of each side is $4x^2y^4$

5 0

3 years ago

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

Tara makes 30 cups of donut topping by mixing sugar and cinnamon. The ratio of sugar to cinnamon is 3:2

AveGali [126]

Answer:

Step-by-step explanation:

3:2 means 3/2 or 3÷2

but its better to leave it as

3/2

4 0

3 years ago

Erica measures the mass of a steel ball several different times. If Erica's results are highly precise, this must mean that A. m

zysi [14]

C is the answer you are looking for.

8 0

3 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago