$=(a^2-10a)-(b^2+6b) +16$
$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$
$=(a-5)^2-25-(b+3)^2+9+16$
$=(a-5)^2-(b+3)^2$
59 is a prime number and an odd number, hope this helps!
Answer:
Step-by-step explanation:
tto blurrry
Answer:
Step-by-step explanation:
Here we are given with two coordinates and asked to determine the distance between them.
Here we are going to use the distance formula, which is given as under
Where
Replacing these values in the distance formula
Hence the Distance is 7.071 units
After 10 years, there will have been 120 deposits. The last one earns 12%/12 = 1% interest, so is mutipied by 1.01. The one before is multiplied by 1.01². Overall, you have the sum of a geometric series of 120 terms with first term 10.10 and common ratio 1.01. That sum is given by the general formula
Sn = a1·(r^n -1)/(r -1)
S120 = 10.10(1.01^120 -1)/(1.01 -1)
S120 = 1010·2.30038689 ≈ 2323
At the end of the 10th year (before the first deposit of the 11th year), the account balance will be
$2,323
