Answer:
The hyperbolas which open horizontally are:
(x+2)^2/3^2-(2y-10)^2/8^2=1
(x-1)^2/6^2-(2y+6)^2/5^2=1
Step-by-step explanation:
A hyperbola with equation of the form:
(x-h)^2/a^2-(y-k)^2/b^2)=1 opens horizontally
Then, the hyperbolas which open horizontally are:
(x+2)^2/3^2-(2y-10)^2/8^2=1
(x-1)^2/6^2-(2y+6)^2/5^2=1
Let x and y be the 2 parts of 15 ==> x + y=15 (given)
Reciprocal of x and y ==> 1/x +1/y ==> 1/x + 1/y = 3/10 (given)
Let's solve 1/x + 1/y = 3/10 . Common denominator = 10.x.y (reduce to same denominator)
==> (10y+10x)/10xy = 3xy/10xy ==> 10x+10y =3xy
But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50
Now we have the sum S of the 2 parts that is S = 15 and
their Product = xy =50
Let's use the quadratic equation for S and P==> X² -SX +P =0
Or X² - 15X + 50=0, Solve for X & you will find:
The 1st part of 15 is 10 & the 2nd part is 5
Answer:
Step-by-step explanation:
fog(x)=f(g(x))
so it comes
<u>so C is the correct answer</u>
Answer:
0.1 seconds
Step-by-step explanation:
10/100=0.1
