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
Answer:
Option (4)
Step-by-step explanation:
Let x = ![\sqrt[4]{7}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B7%7D)
= (x)(x)(x)(x)
= 
= x⁴
Now we substitute the value x = ![\sqrt[4]{7}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B7%7D)
x⁴ = ![(\sqrt[4]{7})^4](https://tex.z-dn.net/?f=%28%5Csqrt%5B4%5D%7B7%7D%29%5E4)
x⁴ = 7
Therefore, Option (4) will be the answer.
Your first step should be to plug in the q and f values they give you. q being 20 and f being 16. From there you want to isolate 1/p by subtracting the 1/20 from both sides. You should be left with 1/p = 1/16 - 1/20. This will come out to be 1/80 or 0.0125. Now that one side is 1/p and the other is 1/80 you can multiply both sides by p in order to get 1=1/80 times p which means that p = 80 . Hope this helps





Now,







Area of rectangle = 38.5 cm².