Answer:
the equation of the hyperbola is y^2/12-x^2/4=1
Step-by-step explanation:
Yes, the number is greater, but there is an explanation. You are dividing a whole number by a decimal less than 1, yes? That means, that the answer you get, when multiplied with the decimal less than 1, will equal 7. And a decimal less than one is really small, so to get the whole number that you are dividing, they will have to multiply with a large number to get that whole number.
Confusing, I know. Let me add an example. Say you're dividing 7 by 0.875. Once divided, your answer is 8. Because 0.875 × 8 equals to 7. The smaller-than-1 decimal will need a larger number to get to the number it is dividing.
7 ÷ 0.875 = 8.
The calculator should help if you want to back this up. Does this make sense, or am I making things more complicated?
Answer:
The first four terms in the sequence 57, 50, 43, and 36
Step-by-step explanation:
Put the numbers in place of x (or n in this case).
You subtract that number you put in place of n by 1.
You then distribute the difference of n and 1.
You then subtract the distributed number by 57
Answer:
(x+4)² + (y-6)² = 81
Step-by-step explanation:
<u>General equation of a circle :</u> (x-h)² + (y-k)² = r²
where (h,k) is at center and r = radius
Here we are given that the circle has a center at (-4,6) and a radius of 9
So we have (h,k) = (-4,6) so h = -4 and k = 6 and radius = 9 so r = 9
<u>Plugging in values into general equation</u>
Recall that the general equation of a circle is (x-h)² + (y-k)² = r²
we have h = -4 , k = 6 and r = 9
==> plug in values
doing so we acquire (x-(-4))² + (y-6)² = 9²
==> apply double negative sign rule
(x+4)² + (y-6)² = 9²
==> evaluate exponent
(x+4)² + (y-6)² = 81
And we are done!
Recall that

Also, recall that for all
,

With
, we can expect
, and with
,
. So from the above identity it follows that

and

and so
