6 · (10 + 2) + (10 − 2) = 80
Answer:
range: (-5, ∞)
domain: (-∞, ∞)
Step-by-step explanation:
given vertex, absolute minimum: (-3, -5)
Range look at y-value
(minimum, maximum) the -5 will be on the left of the comma b/c its the min
If this was a max, you would put the y-value on the right side of the comma
At point of intersection the two equations are equal,
hence, 6x³ =6x²
6x³-6x²=0
6x²(x-1)=0 , the values of x are 0 and 1
The points of intersection are therefore, (0,0) and (1,6)
To find the slopes of the tangents at the points of intersection we find dy/dx
for curve 1, dy/dx=12x, and the other curve dy/dx=18x²
At x=0, dy/dx=12x =0, dy/dx=18x² = 0, hence the angle between the tangents is 0, because the tangents to the two curves have the same slope which is 0 and pass the same point (0,0) origin.
At x=1, dy/dx =12x = 12, dy/dx= 18x² =18, Hence the angle between the two tangents will be given by arctan 18 -arctan 12
= 86.8202 - 85.2364 ≈ 1.5838, because the slope of the lines is equal to tan α where α is the angle of inclination of the line.
Answer:
Step-by-step explanation:
We have to write an equation of a line which passes through the given point (-9,2) and is perpendicular to the given straight line y = 3x - 12 ........... (1)
Now, equation (1) is in the slope-intercept form and the slope of the line is 3.
Let, m is the slope of the required line.
So, 3m = -1
{Since, the product of the slopes of two perpendicular straight lines is -1}
⇒ 
Therefore, the equation of the required line in slope intercept form is
{Where c is a constant}
Now, this above equation passes through the point (-9,2) point.
So, 
⇒ 2 = 3 + c
⇒ c = - 1
Therefore, the equation of the required straight line is (Answer)
<h3>
Answer: 21p^3+5p?-10p-3p'-2</h3><h3>
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Step-by-step explanation:
