At 5: 45 p. m., a jet is located 108 miles due east of a city. a second jet is located due north of the city such that the distance between the two jets is 240 miles.

to the nearest tenth of a mile, what is the distance between the second jet and the city?

214.3 miles.

Step-by-step explanation:

Please find the attachment.

Let x represent the distance between the second jet and the city.

We have been given that at At 5:45 p.m., a jet is located 108 miles due east of a city. A second jet is located due north of the city such that the distance between the two jets is 240 miles.

Upon looking at the attached image we can see that both jets form a right triangle with the city, whose one leg is 108 miles and hypotenuse is 240 miles.  

To solve for x, we will use Pythagoras theorem.

Upon taking square root of both sides of our equation we will get,

Therefore, the distance between the second jet and the city is approximately 214.3 miles.

the science club has 42 people and the robotics club has 33

step-by-step explanation:

a4 - 3a2 - 1

step-by-step explanation:

step   1   :

equation at the end of step   1   :

  ((a4) -   3a2) -   1

step   2   :

trying to factor by splitting the middle term

2.1     factoring   a4-3a2-1  

the first term is,   a4   its coefficient is   1 .

the middle term is,   -3a2   its coefficient is   -3 .

the last term, "the constant", is   -1