Answer: 275 maybe
Step-by-step explanation: i did 46,000 x 0.6% and got 275.
Answer:

Step-by-step explanation:
Your generic line will be
, so you'll need two condition.
Since it has to be perpendicular to
, the product of the slopes needs to be -1, or
, hence your "generic" becomes
. Since it has to pass through [/tex](-5;-4), let's put the coordinates in there: 
your line becomes 
Answer:
A) Gradient = -3
B) 3y - x = 7
Step-by-step explanation:
The curve has the equation;
y = x³ - 6x² + 9x + 1
We are given the pints it passes through as;
A(2,3) and P(3, 1)
A) to find the gradient, we will find the derivative of the given equation.
Thus;
Gradient = y' = 3x² - 12x + 9
At point A, x = 2. Thus;
Gradient = 3(2²) - 12(2) + 9
Gradient = 12 - 24 + 9
Gradient = -3
B) since the gradient of the tangent = -3, it means the gradient of the normal will be; -1/-3 = 1/3
Thus, equation of the normal to the curve at point A will be;
(y - 3) = ⅓(x - 2)
Multiply both sides by 3 to get;
3y - 9 = x - 2
3y - x = 9 - 2
3y - x = 7
Answer:
4x+3y-10=0
Step-by-step explanation:
first find the gradient of the line that is perpendicular to the line of interest.
y2-y1/x2-x1
3--3/6--6
=3+3/6+6
=9/12
3/4
The gradient of the line of interest is the negative reciprocal of 3/4. Therefore m is -4/3
next, substitute information into
y-y1=m(x-x1)
y-2=-4/3 (x-1)
multiply both sides of equation by 3
3y-6=-4 (x-1)
expand brackets
3y-6=-4x+4
rearrange in general form
4x+3y-10=0