Answer:
√
8
≈
3
Explanation:
Note that:
2
2
=
4
<
8
<
9
=
3
2
Hence the (positive) square root of
8
is somewhere between
2
and
3
. Since
8
is much closer to
9
=
3
2
than
4
=
2
2
, we can deduce that the closest integer to the square root is
3
.
We can use this proximity of the square root of
8
to
3
to derive an efficient method for finding approximations.
Consider a quadratic with zeros
3
+
√
8
and
3
−
√
8
:
(
x
−
3
−
√
8
)
(
x
−
3
+
√
8
)
=
(
x
−
3
)
2
−
8
=
x
2
−
6
x
+
1
From this quadratic, we can define a sequence of integers recursively as follows:
⎧
⎪
⎨
⎪
⎩
a
0
=
0
a
1
=
1
a
n
+
2
=
6
a
n
+
1
−
a
n
The first few terms are:
0
,
1
,
6
,
35
,
204
,
1189
,
6930
,
...
The ratio between successive terms will tend very quickly towards
3
+
√
8
.
So:
√
8
≈
6930
1189
−
3
=
3363
1189
≈
2.828427
Answer:
yes a
Step-by-step explanation:
bearing in mind that standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient
![\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{8}{9}}[x-\stackrel{x_1}{(-4)}]\implies y+1=\cfrac{8}{9}(x+4) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{9}}{9(y+1)=9\left( \cfrac{8}{9}(x+4) \right)}\implies 9y+9=8(x+4)\implies 9y+9=8x+32 \\\\\\ 9y=8x+23\implies -8x+9y=23\implies 8x-9y=-23](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Cstackrel%7By_1%7D%7B%28-1%29%7D%3D%5Cstackrel%7Bm%7D%7B%5Ccfrac%7B8%7D%7B9%7D%7D%5Bx-%5Cstackrel%7Bx_1%7D%7B%28-4%29%7D%5D%5Cimplies%20y%2B1%3D%5Ccfrac%7B8%7D%7B9%7D%28x%2B4%29%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B9%7D%7D%7B9%28y%2B1%29%3D9%5Cleft%28%20%5Ccfrac%7B8%7D%7B9%7D%28x%2B4%29%20%5Cright%29%7D%5Cimplies%209y%2B9%3D8%28x%2B4%29%5Cimplies%209y%2B9%3D8x%2B32%20%5C%5C%5C%5C%5C%5C%209y%3D8x%2B23%5Cimplies%20-8x%2B9y%3D23%5Cimplies%208x-9y%3D-23)
Answer:
1.) Exponential Growth
2.) Exponential Decay
3.) Exponential Growth
4.) Exponential Decay
Step-by-step explanation:
<u>1.) </u><u><em>f (x) </em></u><u>= 0.5 (7/3)^</u><u><em>x</em></u>
↓
always increasing
<u>2.) </u><u><em>f (x) </em></u><u>= 0.9 (0.5)^</u><u><em>x</em></u>
<em> </em>↓
always decreasing
<u>3.) </u><u><em>f (x) </em></u><u>= 21 (1/6)^</u><u><em>x</em></u>
↓
always increasing
<u>4.) </u><u><em>f (x) </em></u><u>= 320 (1/6)^</u><u><em>x</em></u>
<em> </em> ↓
always decreasing
<u><em>EXPLANATION:</em></u>
It's exponential growth when the base of our exponential is bigger than 1, which means those numbers get bigger. It's exponential decay when the base of our exponential is in between 1 and 0 and those numbers get smaller.
Yes by 10's and hundreds and ones place and even and odd. And pine and composite
