11
(9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99)
Those numbers divisible by 9 are the multiple of 9; thus need to know how many multiples of 9 there are between 1 and 100:
100 ÷ 9 = 11 r 1 ÷ last multiple of 9 is 11 × 9 (= 99)
→ There are 11 - 1 + 1 = 11 numbers between 1 and 100 which are divisible by 9.
(They are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.)
Step One
Solve for CE as a numerical value
CE = 1/2 * CD = 1/2 66 = 33
CE = 33
Step Two
Solve for x
CD also equals x^2 - 3 which is given.
x^2 - 3 = 33 Add 3 to both sides
x^2 = 33 + 3
x^2 = 36 Take the square root of both sides
sqrt(x^2) = sqrt(36)
x = 6
Step 3
Find the length of AE
AE is given as 6x - 10
x = 6
AE = 6*6 - 10
AE = 36 - 10
AE = 26
Step 4
Find AB
AE = 1/2 AB Definition of bisect
26 = 1/2 AB Multiply by 2
2*26 = AB
AB = 52 <<<<<<<<Answer
The answer is D.
Hopefully this helps! :)
Answer:
the value after 4 years will be 1570.855
Step-by-step explanation:
Hope this helps:)