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:
26.4 m
Step-by-step explanation:
The sine of an angle is the ratio between the opposite side of the angle and the hypotenuse. Hence,
sin 40° = 
= 
c = 
<u> c = 26.4 m (3 sf)</u>
Let 9x^2-1 = y^2
<span>=> 18xdx = 2ydy </span>
<span>=> ydy = 9xdx </span>
<span>lower limit = sqrt(9*2/9 - 1) = sqrt(1) = 1 </span>
<span>upper limit = sqrt(9*4/9 - 1) = sqrt(3) </span>
<span>Int. [sqrt(2)/3,2/3] 1/(x^5(sqrt(9x^2-1)) dx </span>
<span>= Int. [sqrt(2)/3,2/3] xdx/(x^6(sqrt(9x^2-1)) </span>
<span>= 81* Int. [1,sqrt(3)] ydy/((y^2+1)^3y) </span>
<span>=81* Int. [1,sqrt(3)] dy/(y^2+1)^3 </span>
<span>y=tanz </span>
<span>dy = sec^2z dz </span>
<span>=81*Int [pi/4,pi/3] cos^4(z) dz </span>
<span>=81/4*int [pi/4,pi/3] (1+cos(2z))^2 dz </span>
<span>=81/4* Int. [pi/4,pi/3] (1+2cos(2z)+cos^2(2z)) dz </span>
<span>=81/4*(pi/3-pi/4) + 81/4*(sin(2pi/3)-sin(pi/2)) + 81/8 * (pi/3-pi/4) </span>
<span>+ 81/32 *(sin(-pi/3)-sin(pi)) </span>
<span>=81(pi/48+pi/96+1/4*(sqrt(3)/2 - 1) - 1/32 * sqrt(3)/2) </span>
<span>=81/32*(pi+3sqrt(3)-8)</span>
A flat three dimensional solid like a cube, a prism or a pyramid. When you cut out the "net", fold it and glue it together you can see what the three dimensional shape looks like. (It can be used in math)Draw a net of the polyhedron. Calculate the area of each face. Add up the area of all the faces.
Answer:
Step-by-step explanation:
For this, we kind of have to test points to solve it. First, we can test theh interval x<= -2. Plug in -2 and we get a negative value which works. Next, let's try -2<x<0. Plug in -1 and get a positive value which doesn't work.
Next try 0<=x<=7. Plug in any number like 4 and get a negative value which works.
Finally, try x>7. Plug in 8 for example and you would get a positive value. So the solution is
x <= -2 OR 0 <= x <= 7
