Answer:
Step-by-step explanation:
We want to find an equation of a line that's perpendicular to x=1 that also passes through the point (8,-9).
Note that x=1 is a <em>vertical line </em>since x is 1 no matter what y is.
This means that if our new line is perpendicular to the old, then it must be a <em>horizontal line</em>.
So, since we have a horizontal line, then our equation must be our y-value of our point.
Our y-coordinate of our point (8,-9) is -9.
Therefore, our equation is:
And this is in standard form.
And we're done!
Step-by-step explanation:
slope formula when it's perpendicular : m1 × m = -1
Now Differentiate the quadratic function given.
d/dx ( x^2 - x + 1 ) at an x value of -1 from the point "(-1 , 3 )"
= 2x - 1 .... plug in the x value
= 2 × -1 - 1 = -3.
Use this equation...
y - y1 = m(x - x1)
m = slope = derivative at x = -1
y1 = value found by subbing -1 into the original function
x1 = the x value often given.
y - 3 = 1/3 (x + 1)
= 1/3x + 1/3 + 3
y = 1/3x + 10/3.
not sure about this one - it probably intersects at the x & y intercepts of the equation of the normal line.
y = 0 + 10/3 = 10/3
1/3x = 0 - 10/3
x = -10/3 / 1/3 = -10
so the point .... (-10 , 10/3)
Potential roots have the format
factors of the constant term
-------------------------------------
factors of the initial term
Here, 7/3 is the only possible root that has a 7 in the numerator and a factor of 9 in the denominator.
Verify that 7/3 is a root using synthetic division.
Hello,
The initial amount of students is 75. The amount decreased by 80%.
new total = 75-80%
= 15
Thus, 15 students are left.
Faith xoxo
kAnswer: There is a mistake into this i think
Step-by-step explanation:
