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:
because your grandfather has the experience to do work
(2,2)
(x1+x2)/2, (y1-y2)/2
(-2+6)/2=(2)
(-2+6)/2=2
Im giving you instrutions to get the answer its simple:D
The figure above shows a circular sector OAB<span> , subtending an </span>angle<span> of θ radians ... The points A and B lie on the circle so that the </span>angle AOB<span> is 1.8 radians. .... c) </span>Calculate<span> the smallest </span>angle<span> of the </span>triangle<span>ABC , giving the answer in </span>degrees<span>, .... Given that the length of the arc AB is </span>48<span> cm , </span>find<span> the area of the shaded region</span><span>.
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