(1+x^2)^8
=(1+8x^2+8*7/(1*2)x^4+8*7*6/(1*2*3)x^6+8*7*6*5/(1*2*3*4)x^8+....)
=1+8x^2+28x^4+56x^6+70x^8+....)
For x<1, higher power terms diminish in value, hence we can approximate powers of numbers.
1.01=(1+0.1^2) => x=0.1 in the above expansion
(1.01)^8
=1+8(0.1^2)+28(0.1^4)+56(0.1^6) [ limited to four terms, as requested]
=1+0.08+0.0028+0.000056 (+0.00000070)
=1.082856 (approximately)
Answer:

Step-by-step explanation:
The graph of the equation that will contain the points (2, 3) and (3, 2) is the graph that has a slope value that is equivalent to the slope value of the line running through the two points.
Slope of the line running through (2, 3) and (3, 2):
.
Slope (m) = -1.
The equation,
, is given in the slope-intercept form, which means it has a slope value of -1. I.e. the term "-x" is equivalent to -1x. So therefore, the graph of the equation that contains the points (2, 3) and (3, 2) is
.
Answer:
0.07
Step-by-step explanation:
Answer:
$50,000
Step-by-step explanation:
Let x = car sales
We multiply x by 0.02 since she earns 2% along with the added $2500 so that she can end up with $3500
3500 = 0.02x + 2500
Subtract 2500 from both sides
3500 - 2500 = 0.02x + 2500 - 2500
1000 = 0.02x
Divide both sides by 0.02
1000/0.02 = 0.02x/0.02
x = 50000
Answer:
<em>y = 3x - 17</em>
Step-by-step explanation:
Here is the point-slope form of the equation of a line.

If you are given a slope, m, and a point on the line, (x1, y1), you just plug in the values into the equation above, and you get the equation of the line.
In your problem, you are given a point on the line, (5, -2). Now you need the slope of the line. Your line is perpendicular to the given line. The slopes of perpendicular lines are negative reciprocals. If you know the slope of a line, the slope of its perpendicular is found by flipping the fraction and changing the sign.
The given line has slope -1/3.
Flip -1/3 to get -3.
Now change the sign to get 3.
The slope of the line you need is 3. The line passes through point (5, -2).
Now we use the point-slope equation and we plug in the values we have.



