I’m just as confused as you are, what is the question even supposed to be? If they are asking what solution you would use that step in then I think it would be to find the area but I’m not sure
For each of these questions, you need to find the derivative
or
. The slope of the tangent to these curves at the point
is the value of
when
and
. It's also important to know that if the slope of a line is
, then the slope of any line normal/perpendicular to this line is
.
###
![y=\dfrac{6x+3}{3x^2+6x+4}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B6x%2B3%7D%7B3x%5E2%2B6x%2B4%7D)
The derivative is
![y'=\dfrac{(3x^2+6x+4)(6x+3)'-(6x+3)(3x^2+6x+4)'}{(3x^2+6x+4)^2}=\dfrac{6(3x^2+6x+4)-(6x+3)(6x+6)}{(3x^2+6x+4)^2}](https://tex.z-dn.net/?f=y%27%3D%5Cdfrac%7B%283x%5E2%2B6x%2B4%29%286x%2B3%29%27-%286x%2B3%29%283x%5E2%2B6x%2B4%29%27%7D%7B%283x%5E2%2B6x%2B4%29%5E2%7D%3D%5Cdfrac%7B6%283x%5E2%2B6x%2B4%29-%286x%2B3%29%286x%2B6%29%7D%7B%283x%5E2%2B6x%2B4%29%5E2%7D)
![y'=\dfrac{6-18x-18x^2}{(3x^2+6x+4)^2}](https://tex.z-dn.net/?f=y%27%3D%5Cdfrac%7B6-18x-18x%5E2%7D%7B%283x%5E2%2B6x%2B4%29%5E2%7D)
When
, we get a slope of
![y'=\dfrac{6-18-18}{(3+6+4)^2}=-\dfrac{30}{169}](https://tex.z-dn.net/?f=y%27%3D%5Cdfrac%7B6-18-18%7D%7B%283%2B6%2B4%29%5E2%7D%3D-%5Cdfrac%7B30%7D%7B169%7D)
###
![y=x^2+9x+16](https://tex.z-dn.net/?f=y%3Dx%5E2%2B9x%2B16)
The derivative is
![y'=2x+9](https://tex.z-dn.net/?f=y%27%3D2x%2B9)
and so the tangent line at (1, 9) has slope
![y'=2+9=11](https://tex.z-dn.net/?f=y%27%3D2%2B9%3D11)
The line normal to this has slope
. The point-slope and slope-intercept forms of this line are
![y-9=-\dfrac1{11}(x-1)\implies y=-\dfrac x{11}+\dfrac{100}{11}](https://tex.z-dn.net/?f=y-9%3D-%5Cdfrac1%7B11%7D%28x-1%29%5Cimplies%20y%3D-%5Cdfrac%20x%7B11%7D%2B%5Cdfrac%7B100%7D%7B11%7D)
###
![y=9x-14](https://tex.z-dn.net/?f=y%3D9x-14)
The derivative is
![y'=9](https://tex.z-dn.net/?f=y%27%3D9)
so the slope of any line tangent to the curve is 9. The line that passes through (3, 4) is
![y-4=9(x-3)\impleis y=9x-23](https://tex.z-dn.net/?f=y-4%3D9%28x-3%29%5Cimpleis%20y%3D9x-23)
The x-coordinate of the point which divide the line segment is 3.
Given the coordinates in the figure are J(1,-10) and K(9,2) and the 1:3 is the ratio in which the line segment is divided.
When the ratio of the length of a point from both line segments is m:n, the Sectional Formula can be used to get the coordinate of a point that is outside the line.
To find the x-coordinate we will use the formula x=(m/(m+n))(x₂-x₁)+x₁.
Here, m:n=1:3 and x₁=1 from the point J(1,-10) and x₂=9 from the point K(9,2).
Now, we will substitute these values in the formula, we get
x=(1/(1+3))(9-1)+1
x=(1/4)(8)+(1)
x=8/4+1
x=3
Hence, the x-coordinate of the point that divides the directed line segment from k to j into a ratio of 1:3 is 3 units.
Learn about line segments from here brainly.com/question/10240790
#SPJ4
Answer:
45
Explanation:
360 / 8 = 45
Answer:
Amira used 4 balloons for each balloon animal.
Step-by-step explanation:
From the table, we can conclude that:
For 20 animals being sold by her, the leftover balloons were 180.
Then there is an increase in the number of animals being sold. She now sells 29 animals. So, the increase in animals being sold is given as:
29 - 20 = 9.
So, 9 more animals were sold. The balloons used for these 9 animals can be obtained by taking the difference of the leftover balloons for the two days.
The difference of the leftover balloons = 180 - 144 = 36.
So, 36 balloons were used for selling 9 more animals.
Similarly, there is an increase of another 9 animals when 38 animals were sold as 38 - 29 = 9. Also, the balloons used in this case is 36 only as the difference of the leftover balloons is 36.
144 - 108 = 36
Therefore, we conclude that for selling 9 balloon animals, Amira used 36 balloons.
Number of balloons used for 9 balloon animals = 36
∴ Number of balloons used for 1 balloon animal =
(Unitary method)
Thus, Amira used 4 balloons for each balloon animal.