<span>So
I only have ones, tens and hundreds block. I need to find a way how can I fit
the given number 1 718.
=> Since I have hundreds block, I can divide 1 718 by 100
=> 1 718 / 100, how many 100s are there in 1 718
=> 17 hundreds
=> now we only have 18 left
=> 18 / 10, how many 10s are there in 18
=> 1 tens
=> Now we only have 8 left
=> 8 / 1, how many ones are there in 8
=> 8
=> 17 hundreds, 1 tens, and 8 ones.</span><span>
</span>
Let
x--------> the number
we know that
11²=121
12²=144
then
x² must be greater than 121 and less than 144
case a) <span>√115
if x=</span><span>√115
then
x</span>²=115-------> is not greater than 121
case b) <span>√121
</span>if x=√121
then
x²=121-------> is not greater than 121
case c) <span>√136
</span>if x=√136
then
x²=136-------> is greater than 121 and is less than 144------> is ok
case d) <span>√150
</span>if x=√150
then
x²=150------> is not less than 144
therefore
the answer is
√136
Depending on the first term, the sequences
would work.
If the sequence began at 6,
If the sequence began at 1,
If the sequence began at 3,
Answer:
Simply put, algebra is about finding the unknown or putting real life variables into equations and then solving them. ... Algebra can include real numbers, complex numbers, matrices, vectors, and many more forms of mathematic representation
Step-by-step explanation:
For the answer to the question above,
<span>V(n) = a * b^n, where V(n) shows the value of boat after n years.
V(0) = 3500
V(2) = 2000
n = 0
V(0) = a * b^0 = 3500
a = 3500
V(2) = a * b^2
2000 = 3500 * b^2
b = sqrt (2000/3500)
b ≈ 0.76
V(n) = 3500 * 0.76^n
We can check it for n = 1 which is close to 2500 in the graph:
V(1) = 3500 * (0.76)^1
V(1) = 2660
And in the graph we have V(3) ≈ 1500,
V(n) = 3500 * (0.76)^3 ≈ 1536
Now n = 9.5
V(9.5) = 3500 * (0.76)^(9.5)
V(9.5) ≈ 258</span>
