Find the inside angle B of the smaller triangle using the base (4) and height (6)
smaller part of B = arctan(4/6) = 33.69 degrees
full B = 90 degrees - smaller B gives you the other part of B for the left hand side
so 90 - 33.69 = 56.31
now AD = 6 x tan(56.31) = 9 feet
answer is G
<u>Answer-</u>

<u>Solution-</u>
Here, n represents the number of months and f(n) represents the number of laptops in the store after n months.
As the store had 150 laptops in the month of January or at the beginning.
So 
Every month, 20% of the laptops were sold and 10 new laptops were stocked in the store.
As 20% of laptops were sold, so 80% were in the store.
So, after one month total number of laptops in the store,


Again after one month total number of laptops in the store,


Analyzing the pattern, the recursive function f(n) will be,

Answer:
Step-by-step explanation:
There's no answer here. Because -6<21.
The possible digits are:
5, 6, 7, 8 and
9. Let's mark the case when the locker code begins with a prime number as
A and the case when <span>the locker code is an odd number as
B. We have
5 different digits in total,
2 of which are prime (
5 and
7).
First propability:
</span>

<span>
By knowing that digits don't repeat we can say that code is an odd number in case it ends with
5, 7 or
9 (three of five digits).
Second probability:
</span>
Answer:
The answers would be 9 and 1.
Step-by-step explanation:
x= 5±√16 can be separated into two equations.
x=5+√16 and x=5-√16
1. 5+√16 can be solved by simplify √16, which is 4. So then it would become 5+4 which equals 9.
2. 5-√16 can be solved by simplifying √16 which is 4. So then it would become 5-4 which equals 1.
So the values of x are 9 and 1.