Proof:-
In ∆XYZ and ∆VWZ

Hence
∆XYZ
∆VWZ(Side-Angle-Side)
a. The center of the circle will be the coordinate point of the marked point inside the circle.
center: (3,-5)
The radius is the distance between the center and any point on the circle.
radius: 4 units
b. standard form for circles: (x-h)²+(y-k)²=r² where (h,k) is the center
standard form: (x-3)²+(y+5)²=16
c. Simplify equation in standard form to get general form.
(x-3)²+(y+5)²=16
x²-6x+9+y²+10y+25=16
general form: x²+y²-6x+10y+18=0
Hope u find this useful!
Martin charged $10 for every 5 bags
10/5 = 2
Martin charges $2 per bag
He raked 24 bags of leaves
24 x 2 = 48
Martin earned $48 dollars
hope this helps
Answer: 52m cards
Step-by-step explanation:
First part of question says: Karen has m decks of cards.
If Karen has m deck of cards and each deck has 52 cards in it, the total number of cards that all that Karen will have will be 52 multiplied by the number of deck of cards Karen has:
= 52 * m
= 52m
From the given recurrence, it follows that

and so on down to the first term,

(Notice how the exponent on the 2 and the subscript of <em>a</em> in the first term add up to <em>n</em> + 1.)
Denote the remaining sum by <em>S</em> ; then

Multiply both sides by 2 :

Subtract 2<em>S</em> from <em>S</em> to get

So, we end up with
