If the endpoints of a diameter are (6,3) and (2,1) the midpoint is the center of the circle so:
(x,y)=((6+2)/2, (3+1)/2)=(4,2)
Now we need to find the radius....the diameter is:
d^2=(6-2)^2+(3-1)^2
d^2=16+4
d^2=20 since d=2r, r=d/2, and r^2=d^2/4 so
r^2=5
The standard form of the circle is (x-h)^2+(y-k)^2=r^2 and we know:
(h,k)=(4,2) from earlier so:
(x-4)^2+(y-2)^2=5
Let
<span>p ----------------> is the perimeter of the base
</span><span>h ---------------> is the height
</span><span>BA-------------> is the area of the bases
</span><span>LA--------------> is the lateral area
we know that surface area is </span><span>the sum of base areas plus lateral areas
</span><span>then
SA=[BA]+P*h
but remember that
LA=P*h
then
SA=</span>[BA]+LA----------> SA=BA+LA
the answer are the options B.) SA=BA+LA and the option D.) SA=BA+ph
Answer:
see explanation
Step-by-step explanation:
Using the Pythagorean identity
cos²A + sin²A = 1 ( divide terms by cos²A )
+ 
= 
, that is
1 + tan²A = sec²A ← as required
<h3>Answer:</h3>
A) ∠A = ∠A' = 38° and ∠B = ∠B' = 42°
<h3>Explanation:</h3>
The sum of angles in ∆ABC is 180°, so ...
... (2x -2) + (2x +2) + (5x) = 180
... 9x = 180
... x = 20
and the angles of ∆ABC are ∠A = 38°, ∠B = 42°, ∠C = 100°.
___
The sum of angles of ∆A'B'C' is 180°, so ...
... (58 -x) +(3x -18) +(120 -x) = 180
... x +160 = 180
... x = 20
and ∠A' = 38°, ∠B' = 42°, ∠C' = 100°.
_____
The values of angle measures of ∆ABC match those of ∆A'B'C', so we can conclude ...
... A) ∠A = ∠A' = 38° and ∠B = ∠B' = 42°
