Find the negative reciprocal of the slope of the original line and use the point-slope formula
Y -y1 =m(x -x1 ) to find the line perpendicular to y = 3x -2
Answer:
y
2
Step-by-step explanation:
1. switch sides
2. distribute
3. simply (combine like terms) in this case it would go to 22y
4. add 48 to both sides
5. simplify
6. divide
y
2
Answer:
$184, assuming attendance of one course (not class).
Step-by-step explanation:
Total cost for one year (x) = $100 + ($7/ month)x, where x is the number of months.
As there are 12 months in one year:
Total cost (12) = $100 + ($7/monthly)(12) = $184
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
Answer:
yes, they are proportional
Step-by-step explanation:
A proportion is a relation that can be modeled by a straight line through the origin. The equation will have the form ...
y = kx
where x is the independent variable, y is the dependent variable, and k is a constant of proportionality.
__
If you start with the equation relating distance, rate, and time:
d = rt
and you fix the time at 50 seconds, then the equation becomes ...
d = 50t
This is the equation of a proportion with a constant of proportionality of 50. It tells you the distance run is proportional to the rate you run. When this equation is graphed, it is a straight line through the origin.